$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(4(0)y^{2} - 17(0)^{2}y + 12(0)^{3})}{(12(0)y^{2} + 18(0)^{2}y - (0)^{3} + 3y^{3})}=\frac{0}{3y^{3}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(4x(x)^{2} - 17x^{2}(x) + 12x^{3})}{(12x(x)^{2} + 18x^{2}(x) - x^{3} + 3(x)^{3})}=-\frac{1}{32}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(15(0)^{5})}{(18(0)^{3}y^{2} + 8(0)^{4}y + 4(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(15x^{5})}{(18x^{3}(x)^{2} + 8x^{4}(x) + 4x^{5} + 3(x)^{5})}=\frac{5}{11}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(3(0)^{2}y)}{(8(0)y^{2} - 2(0)^{3} + 2y^{3})}=\frac{0}{2y^{3}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{2y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(3x^{2}(x))}{(8x(x)^{2} - 2x^{3} + 2(x)^{3})}=\frac{3}{8}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{((0)^{2}y^{4} - 20(0)^{3}y^{3})}{(12(0)^{4}y^{2} - 14(0)y^{5} + 3(0)^{6} + 2y^{6})}=\frac{0}{2y^{6}}\\&lim_{(0,y)\rightarrow(0+,0-)}\frac{0}{2y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(x^{2}(x)^{4} - 20x^{3}(x)^{3})}{(12x^{4}(x)^{2} - 14x(x)^{5} + 3x^{6} + 2(x)^{6})}=-\frac{19}{3}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(7(0)^{2}y^{4} + 4(0)^{5}y - (0)^{6})}{(5(0)^{2}y^{4} - (0)^{4}y^{2} - 20(0)y^{5} + 2(0)^{6} + 5y^{6})}=\frac{0}{5y^{6}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(7x^{2}(x)^{4} + 4x^{5}(x) - x^{6})}{(5x^{2}(x)^{4} - x^{4}(x)^{2} - 20x(x)^{5} + 2x^{6} + 5(x)^{6})}=-\frac{10}{9}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(4(0)^{3}y^{2} + 7(0)y^{4} - 11(0)^{4}y)}{(2(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(4x^{3}(x)^{2} + 7x(x)^{4} - 11x^{4}(x))}{(2x^{5} + 3(x)^{5})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(19(0)^{2}y^{3} - 2(0)^{5})}{(7(0)^{2}y^{3} + 13(0)y^{4} + 3(0)^{5} + y^{5})}=\frac{0}{y^{5}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(19x^{2}(x)^{3} - 2x^{5})}{(7x^{2}(x)^{3} + 13x(x)^{4} + 3x^{5} + (x)^{5})}=\frac{17}{24}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(6(0)y^{2} - 13(0)^{2}y + 5(0)^{3})}{((0)y^{2} - 20(0)^{2}y + 9(0)^{3} - 3y^{3})}=\frac{0}{3y^{3}}\\&lim_{(0,y)\rightarrow(0+,0+)}\frac{0}{3y^{3}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0+,0+)}\frac{(6x(x)^{2} - 13x^{2}(x) + 5x^{3})}{(x(x)^{2} - 20x^{2}(x) + 9x^{3} - 3(x)^{3})}=\frac{2}{13}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(13(0)^{2}y^{4} - 19(0)^{5}y + 6(0)^{6})}{(2(0)^{6} + 3y^{6})}=\frac{0}{3y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{3y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}-\frac{(13x^{2}(x)^{4} - 19x^{5}(x) + 6x^{6})}{(2x^{6} + 3(x)^{6})}=0\\&\text{Since the two limits are equal, it's possible the limit exists.}\\&\text{Testing additional lines could prove or disprove this.}\end{align*}$

$\displaystyle \begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the x-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(16(0)^{3}y^{3} - 15(0)^{4}y^{2})}{(12(0)y^{5} - 14(0)^{2}y^{4} + (0)^{6} + 5y^{6})}=\frac{0}{5y^{6}}\\&lim_{(0,y)\rightarrow(0-,0-)}\frac{0}{5y^{6}}=0\\&\text{We can make a similar substitution for }y=x:\\&lim_{(x,x)\rightarrow(0-,0-)}\frac{(16x^{3}(x)^{3} - 15x^{4}(x)^{2})}{(12x(x)^{5} - 14x^{2}(x)^{4} + x^{6} + 5(x)^{6})}=\frac{1}{4}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$