is $f(x,y)=(\sqrt[3]{x} + \sqrt[3]{y})^3$ Differentiable at $P=(0,0)$ $f(x,y)=(\sqrt[3]{x} + \sqrt[3]{y})^3$ and let $P=(0,0)$. the partial derivative  at $P$ are $$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(0+h, 0)-f(0,0)}{h}=\frac{(\sqrt[3]{h})^3}{h}=1$$ and $$\frac{\partial f}{\partial y}(x,y)=\lim_{h\to0}\frac{f(0, 0+h)-f(0,0)}{h}=\frac{(\sqrt[3]{h})^3}{h}=1$$ and thus both partial derivative are continuous at $P$ and $f$ is differentiable. But when I derivate directly I get $$\frac{\partial f}{\partial x}(x,y)=\frac{(\sqrt[3]{x} + \sqrt[3]{y})^2}{\sqrt[3]{x^2}}$$ and $$\frac{\partial f}{\partial y}(x,y)=\frac{(\sqrt[3]{x} + \sqrt[3]{y})^2}{\sqrt[3]{y^2}}$$ which are undefined for $P=(0, 0)$ and thus clearly not continuous at $P$. So I'm not sure which derivative is correct and whether $f$ is differentiable or not?!?

For $f$ to be differentiable at $(0,0)$ there must be a linear form $ax+by$
such that
$$f(x,y)-(ax+by)$$
tends to zero quicker than $\sqrt{x^2+y^2}$ as $(x,y)\to 0$, that is
$$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}\to0.$$
The numbers $a$ and $b$ must be the partial derivatives of $f$ at $(0,0)$, that is $a=1$
and $b=1$ so that
$$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}
=\frac{f(x,y)-x-y}{\sqrt{x^2+y^2}}
=3\frac{\sqrt[3]{x^2}\sqrt[3]y+\sqrt[3]{x}\sqrt[3]{y^2}}{\sqrt{x^2+y^2}}
.$$
But if we take $(x,y)=(t,t)$ we get
$$\frac{f(x,y)-(ax+by)}{\sqrt{x^2+y^2}}
=\frac{6t}{\sqrt{2}|t|}=\pm3\sqrt2$$
which certainly does not tend to zero as $t\to0$.
Therefore $f$ is not differentiable at $(0,0)$.
