Is this function can be differentiable at (0, 0)? Let we have $$f(x, y) = \left\{\begin{matrix}
\log(1 + x\sin\sqrt[3]{\frac{y^4}{x}}) & x \neq 0\\
0 & x = 0
\end{matrix}\right.$$
I wanna know if this fuction can be differentiated in $(0, 0)$.
Here's what I have:
$f'_x = f'_y = 0$ cuz $f(x, 0) = f(0, y) = 0$
In this case we only need to prove that $f(x, y) = o(||h||)$ for $h \to 0$
$$\lim_{(x, y) \to (0, 0)} \frac{\log(1 + x\sin\sqrt[3]{\frac{y^4}{x}})}{\sqrt{x^2 + y^2}} = 0$$
But how can I find this limit?

We have that by squeeze theorem
$$\left|x\sin\sqrt[3]{\frac{y^4}{x}}\right| \le |x| \to 0$$
therefore
$$ \frac{\log\left(1 + x\sin\sqrt[3]{\frac{y^4}{x}}\right)}{\sqrt{x^2 + y^2}}=\frac{\log\left(1 + x\sin\sqrt[3]{\frac{y^4}{x}}\right)}{ x\sin\sqrt[3]{\frac{y^4}{x}}}\frac{x\sin\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}}$$
with $\frac{\log\left(1 + x\sin\sqrt[3]{\frac{y^4}{x}}\right)}{ x\sin\sqrt[3]{\frac{y^4}{x}}}\to 1$ therefore all boils down in the following

$$\lim_{(x, y) \to (0, 0)} \frac{x\sin\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}}=0$$

indeed assuming wlog $x>0$ we have
$$\frac{x\sin\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}} \le \frac{x\sqrt[3]{\frac{y^4}{x}}}{\sqrt{x^2 + y^2}}=\frac{\sqrt[3]{y^4x^2}}{\sqrt{x^2 + y^2}}\to 0$$
indeed by polar coordinates
$$\frac{\sqrt[3]{y^4x^2}}{\sqrt{x^2 + y^2}}=\rho \sqrt[3]{\sin^4 \theta \cos^2 \theta} \to 0$$
