Step 1 :First, we substitute $x=0$ into the equation, we get $\lim _{y \rightarrow 0} \frac{y^{3}}{y^{2}}=\lim _{y \rightarrow 0} y=0$
Step 2 :Then, we substitute $y=0$ into the equation, we get $\lim _{x \rightarrow 0} \frac{0}{x^{2}}=0$
Step 3 :Next, we substitute $y=4x$ into the equation, we get $\lim _{x \rightarrow 0} \frac{4x^{2}+64x^{3}}{16x^{2}+x^{2}}=\lim _{x \rightarrow 0} \frac{4x+64x^{2}}{17x}=\lim _{x \rightarrow 0} \frac{4+64x}{17}=\frac{4}{17}$
Step 4 :Since the limits along the paths $x=0$, $y=0$ and $y=4x$ are not the same, the limit does not exist
Step 5 :So, the final answer is \(\boxed{\text{DNE}}\)