Given $f(x,y)= \left\{ \begin{array}{ll} \frac{xy^{2}}{x^{2}+y^{2}} & x^2+y^2> 0 \\ 0 & x=y=0 \end{array} \right.$
and $x(t)=y(t)=t$ and computation of $\frac{\mathrm{d} }{\mathrm{d} t} \hspace{1mm} f(x(t),y(t))))|_{t=0}$ is $\frac{1}{2}$ but $f'_{x}(0,0)=f'_{y}(0,0)=0$. In this case the chain rule computation is not coinciding with the direct computation.
Is there a property or a necessary condition of the function, $f(x,y)$ that breaks the chain rule in the above case?
I have tried to approach this formulation by the definition of chain rule. Instead of directly plugging the value of t, the composite function can be differentiated with respect to the independent variable t as:
$\frac{\mathrm{d} }{\mathrm{d} t} \hspace{1mm} f(x,y)|_{t=0}= f'_{x}* \frac{\mathrm{d} }{\mathrm{d} t} \hspace{1mm} x(t) + f'_{y}* \frac{\mathrm{d} }{\mathrm{d} t} \hspace{1mm} y(t) $.
When $t=0, x=y=0 \hspace{1mm}, f'_{x}=f'_{y}=0$ then $\frac{\mathrm{d} }{\mathrm{d} t} \hspace{1mm} f(x,y)|_{t=0}=0$ after plugging the following values.
But I am not getting that why on directly substituting the value of t in the above function yields incorrect results.
Please let me know if my approach to the above formulation is correct or does chain rule works on the concept that the functions should be continuous and differentiable at the given point?