Vtyjkyuk

Define $f:mathbbR^2rightarrow mathbbR$ by $f(x,y)=displaystyle fracy^3-sin^3xx^2+y^2$ if $(x,y)neq (0,0)$ and $f(0,0)=0$.
My question is, is $f$ differentiable at $(0,0)$?

First I guessed it is not. So, I tried to prove that $f$ is not continuous at $(0,0)$ by showing the limits are different for two different paths approaching $(0,0)$. But could not find it easily. Then I tried proving it is differentiable.

So I took $displaystyle lim_(x,y)rightarrow (0,0)fracf(x,y)-f(0,0)-nabla f(0,0).(x,y)sqrtx^2+y^2=lim_(x,y)rightarrow (0,0)fracx^3-sin^3x(x^2+y^2)^(3/2)$. But failed to prove that this limit is $0$.

So, how could I determine whether $f$ is differentiable at $(0,0)$?

Your approach is completely correct, but the execution is wrong. First of all, the gradient at the origin is not $(0,0)$; it should be $(-1,1)$. Look carefully at $f(x,0)$, and you'll see that
$$lim_hto 0 fracf(h,0)-f(0,0)h = -1,$$
etc.

So we need to decide if
$$lim_(x,y)to (0,0)fracfracy^3-sin^3xx^2+y^2 - (-x+y)sqrtx^2+y^2 = 0$$
or not. I'll let you work out the algebra, but you should need to determine whether
$$lim_(x,y)to (0,0) fracxy(x-y)(x^2+y^2)^3/2 = 0;$$
what do you think?

So, because $nabla f(0,0) = (-1,1)$you need to check if $$lim_(x,y) to (0,0) fracx^3 - (sin(x))^3 + xy(y-x)(x^2 + y^2)^frac32 = 0$$
We can take $(x,y) to (0,0)$ along any path, so choose convergence along the positive $x-$axis, i.e fix $y=0$ and let $x to 0+$
We get $lim_xto 0+ -(fracsin(x)x)^3 = 0$

Now, set $y = kx$ and let $ x to 0+$ The limit becomes

$lim_yto 0+ fracx^3 - sin^3(x)+ x^3 (k^2 - k)x^3(1 + k^2)^frac32 neq 0$ when (say) $k = 0.5$
So, the limit does not exist and the function is not differentiable.