Answer
The partial derivatives exist at $( 0,0)$
$f$ is not continuous and not differentiable at $( 0,0)$
Work Step by Step
Given
$$f(x, y)=\left\{\begin{array}{ll}{\frac{3 x^{2} y}{x^{4}+y^{2}},} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$
Firstly:
\begin{array}{c}{f_{x}(0,0)=\lim _{\Delta x \rightarrow 0} \frac{f(\Delta x, 0)-f(0,0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\frac{0}{(\Delta x)^{4}}-0}{\Delta x}=0} \\ {f_{y}(0,0)=\lim _{\Delta y \rightarrow 0} \frac{f(0, \Delta y)-f(0,0)}{\Delta y}=\lim _{\Delta y \rightarrow 0} \frac{\frac{0}{(\Delta y)^{2}}-0}{\Delta y}=0}\end{array}
So, the partial derivatives exist at $( 0,0)$
Secondly:
\begin{align}{}{\text { Along the line } y=x:\\
\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{x \rightarrow 0} \frac{3 x^{3}}{x^{4}+x^{2}}=\lim _{x \rightarrow 0} \frac{3 x}{x^{2}+1}=0} \\ {\text { Along the curve } y=x^{2} : \\
\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\frac{3 x^{4}}{2 x^{4}}=\frac{3}{2}}\end{align}
this implies that $f$ is not continuous at $(0, 0)$ . So, $f$ is not differentiable at $( 0,0)$
