Let $f:\mathbb{R}^2 \longrightarrow \mathbb{R}$
$f(x,y)^t:= \begin{cases} \frac{x^2y}{x^6+2y^2} &, (x,y)^t\neq (0,0)^t \\ 0 &, (x,y)^t=(0,0)^t \end{cases}$
($(.)^t$ means transponed)
I need to determine the partial derivatives of $f$ and decide where the derivatives are continuous.
So for $(x,y)^t \neq (0,0)^t$:
$\frac{\partial f}{\partial x} = \frac{-4x^7y+4xy^3}{(x^6+2y^2)^2}$
$\frac{\partial f}{\partial y} = \frac{x^8-2x^2y^2}{(x^6+2y^2)^2}$
And for $(x,y)^t = (0,0)^t$:
$\lim_{s\rightarrow 0}\frac{1}{s}(f(s,0)^t-f(0,0)^t)=0=\lim_{s\rightarrow 0}\frac{1}{s}(f(0,s)^t-f(0,0)^t)$.
Now let $(x_n,y_n)^t \in \mathbb{R}^2 \neq (0,0)^t$ with $\lim_{n\rightarrow \infty}(x_n,y_n)^t=(0,0)^t$. I need to check if $\frac{-4x^7y+4xy^3}{(x^6+2y^2)^2}$ with $x:=x_n, y:=y_n$ converges to $0$ and the same for $\frac{x^8-2x^2y^2}{(x^6+2y^2)^2}$ which they don't according to maple.
But how do I prove this?
I thought about defining $x_n,x'_n$ and $y_n,y'_n$ which all converge to zero but $\frac{-4x_n^7y_n+4x_ny_n^3}{(x_n^6+2y_n^2)^2} \neq$ the same with ${x'}_n$ instead of $x_n$
You can easily show with the definition that $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$
Now we need to check if $$\lim_{(x,y) \to (0,0)}\frac{-4x^7y+4xy^3}{(x^6+2y^2)^2} \neq 0$$
By taking polar coordinates: $$x=r \cos{t}$$ $$y=r\sin{t}$$ we have that:
$$\lim_{r \to 0} \frac{4r^8 \cos^7{t}\sin{t}+4r^4 \cos{t} \sin^3{t}}{(r^6\cos^6{t}+2r^2\sin^2{t})^2}=\lim_{r \to 0} \frac{4r^4 \cos^7{t}\sin{t}+4 \cos{t} \sin^3{t}}{(r^4\cos^6{t}+2\sin^2{t})^2}$$
You can see that for this last limitwe gain different values for different values of $t$ (for example $t=0,\frac{\pi}{4}$)thus the limit does not exist.
So the first partial derivative of $f$ is not continuous at the origin.
Apply the same method for the other partial derivative to check its continuity.