Examine where $\nabla f = (0, 0)$ 
Given $f: \Bbb R^2 \rightarrow \Bbb R$ defined by
$$f(x) = \|x\|^4 - \|x\|^2$$
where $x = (x_1, x_2) \in \Bbb R^2$,  I have to examine where $\nabla f = (0, 0)$.

Approach
Since $||x||$ was not defined, I can only assume that this is the euclidian norm, which means that,
$||x||^4 := (x_1)^2 + (x_2)^2,$
$||x||^2 := x_1 + x_2.$
Therefore,
$df \over dx_1$ = $2x_1,$
$df \over dx_2$ =$ 2x_2,$
such that,
$\nabla f = (2x_1, 2x_2) = (0, 0)$
only holds for $(x_1, x_2) = (0, 0)$.
I have to admit that this seems a little bit too easy. Did I make a mistake when I defined $||x||^4$ and $||x||^2$?
 A: First we have :
$$
f(x) = ((x_1)^2 +(x_2)^2)^2 - (x_1^2 + x_2^2) = x_1^4 + 2x_1^2x_2^2 + x_2^4 -x_1^2 -x_2^2
$$
Then we have
$$
\frac{\partial}{\partial x_1} f(x) = 4x_1^3 + 4x_1 x_2^2 - 2x_1 = 0
$$
$$
\frac{\partial}{\partial x_2} f(x) = 4x_2^3 + 4x_1^2 x_2 - 2x_2 = 0
$$
Clearly $(0,0)$ is a solution. We can then divide the first equation by $x_1$ and the second by $x_2$ :
$$
4x_1^2 + 4x_2^2 - 2 = 0\\
4x_2^2 + 4x_1^1 - 2 = 0
$$
We see that both equations are actually equivalent : $x_1^2 + x_2^2  = 1/2$
which is the equation of the circle centered at $(0,0)$ of radius $\sqrt{1/2}$.
Therefore the gradient is $0$ at the origin and on this circle.
A: Assuming that $\|\cdot\|$ denotes the $2$-norm,
$$f (x) = \|x\|_2^4 - \|x\|_2^2 = \|x\|_2^2 \, (\|x\|_2^2 - 1) = (x^T x) \cdot (x^T x - 1)$$
Taking the gradient,
$$\nabla f (x) = 2 (\|x\|_2^2 - 1) \, x + 2 \|x\|_2^2 \, x = 2 \, (2 \|x\|_2^2 - 1) \, x$$
Hence, the gradient vanishes at the origin, $x = 0_2$, and on the circle $\|x\|_2 = \frac{1}{\sqrt 2}$.
