Determine that this multivariable function is differentiable 
$$
f(x,y) = \begin{cases}
0, & x = 0 = y \\
\frac{x^3}{x^2 + y^2}, & \text{otherwise}
\end{cases}
$$

So, I did determine that this function is continuous(correct me if I'm wrong here). But, the issue is that the first order partial derivative with respect to $y$ is not continuous(I found this out by taking the partial derivative away from the origin and then switching to polar coordinates and taking the limit as $r \rightarrow 0$. So, I cannot use the differentiability theorem(existence of continuous first order partial derivatives guarantee differentiability). So, what do I do from here?
 A: My answer shows it is not differentiable at origin. Kindly check whether  I have  done any calculation error.
$${f_x} (0,0)=\mathop {\lim }\limits_{x \to 0} {{f(x,0) - f(0,0)} \over x} = \mathop {\lim }\limits_{x \to 0} {{x-0} \over x} = 1$$
$${f_y} (0,0)=\mathop {\lim }\limits_{y \to 0} {{f(0,y) - f(0,0)} \over y} = \mathop {\lim }\limits_{y \to 0} {{0 - 0} \over y}=0.$$
Therefore
$$I=\lim_{(x,y)\rightarrow (0,0) } {{f(x,y) - f_x(0,0)x-f_y(0,0)y-f(0,0)} \over{\sqrt{x^2+y^2}}} = \lim_{(x,y)\rightarrow (0,0) } {{{x^3/(x^2+y^2)}-x-0-0}\over {\sqrt{x^2+y^2}}}$$
$$=\lim_{(x,y)\rightarrow (0,0) } {-xy^2\over{(x^2+y^2)\sqrt{x^2+y^2}}}. $$ Calculate the limit value of the second using polar co-ordinates, by taking  $x=r\cos\theta$ and $y=r\sin\theta$ and $r\rightarrow 0.$
Then $I=\lim_{r\rightarrow 0}\cos \theta\sin^2\theta =f(\theta).$ Limit value depends on the path. So limit does not exist and the function is not differentiable at the origin.
A: $∆f=f(0+h,0+k)-f(0,0)$
$= \frac{h^3}{h^2+k^2}$
And $f_x = 1$, $f_y=0$.
Therefore $df = hf_x+kf_y=h$.
If the function is differentiable then $\lim_{(h,k)\to (0,0)} \frac{∆f-df}{\sqrt{h^2+k^2}}$ must be $0$.
Now, $\frac{∆f-df}{\sqrt{h
^2+k^2}}= -\frac{hk^2}{\sqrt{h^2+k^2}(h^2+k^2)}....(i)$
Putting $h=r cos\theta, k=rsin\theta$  in $...(i)$ we get,
$\frac{∆f-df}{\sqrt{h^2+k^2}}$
$=-cos\theta sin\theta$
Which in general doesn't approach zero as $\sqrt{h^2+k^2}=r \to 0$.
Hence $f(x,y)$ isn't differentiable at $(0,0)$.
