We're given :
$$ f(x,y) = \begin{cases} (x^{2}+y)\sin(\dfrac{1}{x^{2}+y^{2}}) & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \\ \end{cases} $$
We need to show that $f_x(0,0)=0$.
I found out the partial derivative as :
$f_x(x,y) = (2x)\sin(\dfrac{1}{x^{2}+y^{2}}) + (x^{2}+y)\cos(\dfrac{1}{x^{2}+y^{2}})(\dfrac{-2x}{x^{2}+y^{2}})$.
But putting $(x,y)=(0,0)$ here , doesn't yield any result..
Could anyone guide me through this problem ?
By definition, $$ f_x(x,y) = \lim\limits_{h\rightarrow 0}{\frac{f(x+h,y)-f(x,y)}{h}}.$$ In this case, we have $$ f_x(0,0) = \lim\limits_{h\rightarrow 0}{\frac{(h^2+0)\sin\left(\frac{1}{h^2+0^2}\right) - 0}{h}}.$$ Take it from there...