Differentiability of piecewise functions Check whether the function is differentiable:
$$f:\mathbb{R}^2\rightarrow \mathbb{R}$$
$$f= \begin{cases} 
      \frac{x^3-y^3}{x^2+y^2} & (x,y)\neq (0,0) \\
      0 & (x,y) = (0,0) \\
   \end{cases}
$$
So what I did is I calculated the partial derivatives of the function in point $(0,0)$. I got:
$$\frac{∂f}{∂x}\left(0,0\right)=lim_{t\rightarrow 0}\left(\frac{f\left(t,0\right)-f\left(0,0\right)}{t}\right)=lim_{t\rightarrow 0}\left(\frac{t^3}{t^3}\right)=1$$and
$$\frac{∂f}{∂y}\left(0,0\right)=lim_{t\rightarrow 0}\left(\frac{f\left(0,t\right)-f\left(0,0\right)}{t}\right)=lim_{t\rightarrow 0}\left(\frac{-t^3}{t^3}\right)=-1$$
And since the answers I got are not equal, that means the function isn't partially derivable in point $(0,0)$ so it isn't differentiable either?
I'm not sure whether what I did was right, differentiability is still a little unclear to me, for multivariable functions. I also asked about it here Differentiability of function definition but have yet to get an answer. Can someone tell me if I'm on the right track at least?

This is wrong. Being partially differentiable means that the partial derivatives exist, and you have shown this by showing the limits to exist. The partial derivatives need not coincide! To show that $f$ is differentiable a sufficient conditon is that the partial derivatives exist and are continous. To show that $f$ is not differentiable, it suffices to show that the partial derivatives not not exist.
