Differentiability at (0,0).

As you have correctly computed that $\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=0$.

A function is said to be differentiable at a point $(a,b)$ iff there exist $\phi(h,k)$ and $\psi(h,k)$ from $\mathbb{R}^{2}\to\mathbb{R}$ such that

$f(a+h,b+k)=h\frac{\partial f}{\partial x}\vert_{(a,b)}+k\frac{\partial f}{\partial y}\vert_{(a,b)}+h\phi(h,k)+k\psi(h,k)$ such that $\phi(h,k)\to 0\quad \text{and}\quad\psi(h,k)\to 0$ as $(h,k)\to(0,0)$.

So if our given function is differentiable.

Then $f(0+h,0+k)=h\frac{\partial f}{\partial x}+k\frac{\partial f}{\partial y}+h\phi(h,k) +k\psi(h,k)$ where $\psi$ and $\phi$ both $\to 0$ as $(h,k)\to (0,0)$ .

Then $\dfrac{h^3k^3}{h^4+k^4}=h\phi(h,k)+k\psi(h,k)$.

Take $\displaystyle\phi(h,k)=\frac{h^{2}k^{3}}{2(h^{4}+k^{4})}$ and $\displaystyle\psi(h,k)=\frac{k^{2}h^{3}}{2(h^{4}+k^{4})}$.

Then both $\phi$ and $\psi$ tend to $0$ as $(h,k)\to(0,0)$.

Hence you have differentiability.

In a Equivalent formulation of Differentiability.

A function is differentiable at a point $(a,b)$ iff there exist $\psi(h,k)$

$f(a+h,b+k)=h\frac{\partial f}{\partial x}\vert_{(a,b)}+k\frac{\partial f}{\partial y}\vert_{(a,b)}+\sqrt{h^{2}+k^{2}}\cdot \psi(h,k)$ such that $\psi(h,k)\to 0$ as $(h,k)\to(0,0)$.

In this case you just take $\displaystyle\psi=\frac{h^3k^3}{(h^4+k^4)\sqrt{(h^{2}+k^{2})}}$ . Then this would also imply differentiablity.

It's just a matter of what convention you follow. Many people like this norm formulation. I myself prefer the previous one. They are both equivalent.

Lastly, I would say that you should try and use polar coordinates to show that $\psi$ and $\phi$ tend to $0$. It is very easy. $\cos^{4}(\theta)+\sin^{4}(\theta)\geq \frac{1}{2}$ always.