Checking the differentiability of the two variables function Let:
$f(x,y)=\frac{x^3+y^3}{x^2+y^2}$ 
It is required to test the differentiability of the given two variables function at $(0,0)$. The value of function at $(0,0)$ is $0$.
My approach
I conclude that function is not differentiable since:
$f_{x}(0,0)=f_{y}(0,0)=1$ and
$f(x,y)=x+y+x(\frac{-y^2}{x^2+y^2})+y(\frac{-x^2}{x^2+y^2})$. Here, 
$A(x,y)=\frac{-y^2}{x^2+y^2}$ and $B(x,y)=\frac{-x^2}{x^2+y^2}$. These two functions of $x$ and $y$ are not tending towards zero as $(x,y)\rightarrow(0,0)$. Hence, by definition the function is not differentiable. 
I always make mistake in the differentiability test of two variables. I am also not sure whether my conclusion is correct or wrong. So, please help me identify any mistakes.
Thanks.

$\require{cancel}$
$$1.\,\,\,\,\,\partial_xf(0,0)=\lim_\limits{h\to0}\frac{f(0+h,0)-f(0,0)}{h}=\lim_\limits{h\to0}\frac{h^3/h^2}{h}=1.$$
$$2.\,\,\,\,\,\partial_yf(0,0)=\lim_\limits{h\to0}\frac{f(0,0+h)-f(0,0)}{h}=\lim_\limits{h\to0}\frac{h^3/h^2}{h}=1.$$
$$3.\,\,\,\,\,\lim_\limits{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\partial_xf(0,0)x-\partial_yf(0,0)y}{\sqrt{x^2+y^2}}=\lim_\limits{(x,y)\to(0,0)}\frac{\frac{x^3+y^3}{x^2+y^2}-x-y}{\sqrt{x^2+y^2}}=\lim_\limits{(x,y)\to(0,0)}\frac{\cancel{x^3}+\cancel{y^3}-\cancel{x^3}-x^2 y-x y^2-\cancel{y^3}}{\sqrt{(x^2+y^2)^3}}=\lim_\limits{(x,y)\to(0,0)}\frac{-xy\,(x+y)}{\sqrt{(x^2+y^2)^3}}=\lim_\limits{r\to0}\frac{-\cancel{r^3}\cos(\theta)\sin(\theta)\,(\cos(\theta)+\sin(\theta))}{\cancel{r^3}\sqrt{\cancel{(\cos^2(\theta)+\sin^2(\theta))^3}_1}}=\cos(\theta)\sin(\theta)\,(\cos(\theta)+\sin(\theta))=\text{Indeterminate.}$$
Since the last limit is not $0$ then the function is not differentiable at $(0,0)$.
