Consider the funcion $f(x, y)=\displaystyle\frac{2xy^2}{x^2+y^4}$ for $(x, y)\neq(0,0)$ and $f(0, 0)=0$ and the curve $\gamma(t) =(t, t, z(t))$, $t\in\mathbb{R}$, whose image is contained in the graph of $f$. Prove that the Tangent line $T$ in $\gamma(0)$ is not contained in the plan given by $$z-f(0,0)=\frac{\partial f}{\partial x}(0,0)(x-0)+ \frac{\partial f}{\partial y}(0,0)(y-0)$$
In this case I have
$$\frac{\partial f}{\partial x}(0,0)=\displaystyle\lim_{x\rightarrow 0}= \frac{\dfrac{2x0^2}{x^2+0^4}-0}{x-0}=0$$
and
$$\frac{\partial f}{\partial y}(0,0)=\displaystyle\lim_{y\rightarrow 0}= \frac{\dfrac{2\cdot 0y^2}{0^2+y^4}-0}{y-0}=0$$
Moreover $\gamma(0)$ belongs to the plane and
What to do next?
Could anyone please help me
$f(x, y)=\begin{cases}\displaystyle\frac{2xy^2}{x^2+y^4}; (x,y)\ne (0,0)\\ 0; (x,y)=(0,0) \end {cases}$,
is actually the level surface of $g(x,y,z)=f(x,y)-z $.
Now partial derivatives, $\frac{\partial f (0,0)}{\partial x}=0=\frac{\partial f (0,0)}{\partial y}$ as you have rightly shown.
Tangent plane to $g(x,y,z)=0$ at $(0,0,f(0,0))$ is given by:
$\nabla g(0,0,0).(x-0,y-0,z-f(0,0))=0$, where $.$ is dot product and $\nabla $ is gradient of $g$ , which is equal to $(f_x (0,0), f_y(0,0), -1)$ at $(0,0,f(0,0))$
Simplifying the above to get: $z-f(0,0)=\frac{\partial f}{\partial x}(0,0)(x-0)+ \frac{\partial f}{\partial y}(0,0)(y-0)$
$\implies z=0 \tag{1}$
If tangent to $\gamma (0)$ is contained in $(1)$, then we must have $\gamma '(0)\perp \text{Normal to plane in (1)}$, that is we must have $\gamma'(0). (0,0,1) =0\tag {2}$
Now we have, $\gamma'(0)=(1,1,2)$ and clearly this doesn't satisfy $(2)$ and hence tangent to $\gamma (0)$ can't be contained in $(1)$
Response to comment: Why is $\gamma'(0)=(1,1,2)$?
$\gamma (t)=(t,t,z(t))\implies \gamma'(t)=(1,1,z'(t))$
$f(t,t)=\frac{2t}{1+t^2}=z(t)$, where $t\ne 0$ and $z(0)=0$ and hence $z'(0)=\lim_{t\to 0}\frac{z(t)-z(0)}{t}=\lim_{t\to 0} \frac{2}{1+t^2}=2$