Consider the problem minimize $f(x_1,x_2) = (x_2 −x_1^2)(x_2 −2x_1^2)$ (i) Show that the ﬁrst- and second-order necessary conditions for optimality are satisﬁed at $(0,0)^T$.
(ii) Show that the origin is a local minimizer of f along any line passing through the origin (that is, $x_2 = mx_1$).
(iii) Show that the origin is not a local minimizer of f (consider, for example, curves of the form $x_2 = kx_1^2$)
Okay, so for a point to satisfy both the first and second-order necessary conditions for optimality, it must satisfy:
I: gradient of the function at the point $= 0 $(stationary point)
II: hessian matrix of the function at the point must be positive semidefinite.
While the sufficient conditions for optimality are:
I: gradient of the function at the point $= 0 $
II: hessian matrix of the function at the point must be positive definite.
And a point that meets them can be considered a strict local minimum.
Therefore, for the part (i)
I computed the gradient of the function at the point $(x_1,x_2)=(0,0)$ and it is equal to zero.
And the Hessian Matrix:
$\pmatrix{-6·x_2+42·x_1^2&-6·x_1\\ -6·x_1 &2}$ that has eigenvalues $0,2 ≥0$ at the origin and therefore is a Positive Semi-Definite Matrix. 
Therefore, both necessary conditions are met. Part (i) is complete.
For part (ii):
I have to prove the origin as a local minimizer, so I first begin trying to prove the sufficient conditions to see if it then can be stated as a strict local minimizer.
1) Gradient Column: $\pmatrix{2·x_2·m-3·(2·x_1·x_2+x_1^2·m)+8·x_1^3\\ 2·x_2-3·x_1^2}$
Which equals $0$ at the origin (condition 1 met)
2) Hessian Matrix: $\pmatrix{2·m^2-3·(2·x_2+4·x_1·m)+24·x_1^2 &2·m-6·x_1\\ 2·m-6·x_1 &2}$
particularized for the origin: $\pmatrix{2·m^2&2·m\\ 2·m &2}$
And has eigenvalues: $(0,1/(2·m^2+2))$ which are ≥0 so the matrix is PSD (not PD) and we can't therefore conclude that the origin meets the sufficient conditions (and we can't say it is a local minimizer).
What other way can I work through in order to prove the origin as a local minimizer?
For part (iii) I only have to prove the necessary conditions wrong, which is easy, so there are no doubts about that part.
Thanks for the help and time spent reading over this.
Regards!

For part (ii), showing that the origin is a local minimizer of $f$ along any line $x_2 = mx_1$ is basically a one-variable minimization problem, so there is no need to compute a Hessian matrix. You need to show that $g_m(x_1) = f(x_1,mx_1) = (mx_1-x_1^2)(mx_1-2x_1^2)$ has a local minimum at $x_1 = 0$. 
For part (iii), You already showed that the origin does satisfy the necessary conditions for being a minimizer. However, you need to show that despite satisfying the necessary conditions, the origin is NOT a local minimizer of $f$. Hint: for any $t \neq 0$ we have $f(t,\frac{3}{2}t^2) = -\frac{1}{4}t^2 < 0 = f(0,0)$. Based on this, does $f$ attain a local minimum at $(0,0)$?
