Def.(Differentiable and Gradient)
Let $f:S\mapsto\mathbb{R}$ where $S=S^\circ\subseteq \mathbb{R},x\in S$
$$\exists m\in\mathbb{R},s.t.\lim_{h\to0}\frac{f(x+h)-f(x)-m\cdot h}{\Vert h\Vert}=0$$
$$\Leftrightarrow\nabla f(x)=m$$
Then $f$ is differentiable at $x$ with gradient $m$
Let $f:\mathbb{R}^2\mapsto\mathbb{R}$
$$f(x,y)=
\begin{cases}
\frac{y^3-x^8y}{x^6+y^2}& (x,y)\neq(0,0)\\\\
0& (x,y)=(0,0)
\end{cases} $$Is $f$ differentiable at $(0,0)?$
Since for all $x,y\text{ in }\mathbb{R},f(x,0)=0\text{ and }f(0,y)=y$
Then if $\nabla f(0)\text{ exists}$ implies $\nabla f(0)=(0,1)$
That if the following limit exists and equals to $0$, $f$ must be differentiable at $(0,0)$
\begin{align}
& \lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-(0,1)\cdot(x,y)}{\sqrt{x^2+y^2}}\\
& =\lim_{(x,y)\to(0,0)}\frac{f(x,y)-y}{\sqrt{x^2+y^2}} \\
& =\lim_{(x,y)\to(0,0)}\frac{\frac{y^3-x^8y}{x^6+y^2}-y}{\sqrt{x^2+y^2}} \\
& =\lim_{(x,y)\to(0,0)}-\frac{\left(yx^8+yx^6\right)\sqrt{y^2+x^2}}{\left(y^2+x^6\right)\left(y^2+x^2\right)}\\
& =\lim _{\left(x,\:y\right)\to \left(0,\:0\right)}-\frac{y}{\sqrt{y^2+x^2}}\cdot \frac{x^8+x^6}{y^2+x^6}\\
\end{align}
$\text{Update:}$
$\text{Let $x=r\cos(\theta),y=r\sin(\theta)$ we have}$
\begin{align}
& =\lim _{r\to \:0}-\frac{r\sin \left(θ\right)}{\sqrt{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^2}}\cdot \frac{\left(r\cos \left(θ\right)\right)^8+\left(r\cos \left(θ\right)\right)^6}{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^6}\\
& =-\sin \left(θ\right)\cdot \lim _{r\to \:0}\frac{r}{\sqrt{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^2}}\cdot \lim _{r\to \:0}\frac{\left(r\cos \left(θ\right)\right)^8+\left(r\cos \left(θ\right)\right)^6}{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^6}\\
& \neq-\sin \left(θ\right)\cdot\sqrt{\lim _{r\to \:0}\frac{r^2}{\left(r\sin \left(θ\right)\right)^2+\left(r\cos \left(θ\right)\right)^2}}\cdot\lim _{r\to \:0}\frac{r^4\cos ^6\left(θ\right)\left(r^2\cos ^2\left(θ\right)+1\right)}{\sin ^2\left(θ\right)+r^4\cos ^6\left(θ\right)}\\
& =-\sin(\theta)\cdot \text{d.n.e}\cdot0=\text{d.n.e}
\end{align}
Therefore $f$ isn't differentiable at $(0,0).$$\tag*{$\square$}$
Is this calculation correct $?$
Any help would be appreciated.
Best Answer
We have that limit doesn't exist
$$-\frac{\left(yx^8+yx^6\right)\sqrt{y^2+x^2}}{\left(y^2+x^6\right)\left(y^2+x^2\right)} =-\frac{y}{\sqrt{y^2+x^2}}\cdot \frac{x^8+x^6}{y^2+x^6} $$
indeed
$$-\frac{y}{\sqrt{y^2+x^2}}=-\sin \theta$$
and as $y=t$ and $x=t\to 0$
$$\frac{x^8+x^6}{y^2+x^6}=\frac{t^8+t^6}{t^2+t^6}\to 0$$
but as $y=t^3$ and $x=t\to 0$
$$\frac{x^8+x^6}{y^2+x^6}=\frac{t^8+t^6}{2t^6}\to \frac12$$