Show that the restriction of $f$ to any straight line through $B$ has a local minimum in $B$ I have the following problem:

We have the function: $f(x,y)=(x^2+y^2-6x)\cdot (\frac{1}{8}y^2-x)$
We also have the point $B=(0,0)$.
Show that the restriction of $f$ to any straight line through $B$ has a local minimum in $B$. Is this enough to prove that $f$ has a local minimum in $B$? Does $f$ actually have local minimum in $B$?

Okay, so I am having some trouble with this. I have found the partiel derivatives for $f$.
$f'_x=(2x-6) \cdot (\frac{y^2}{8}-x)-x^2-y^2+6x$
$f'_y=2y(\frac{y^2}{8}-x)+\frac{(x^2+y^2-6x)y}{4}$
Inserting $(0,0$ into both $f'_x$ and $f'_y$ yields:
$f'_x(0,0)=0$
$f'_y(0,0)=0$
Here, I just showed that $B$ is a stationary point, but I don't really know how to continue from here, and show that any straight line has a local minimum in $B$.
I hope someone can help me.

Straight lines through $(0, 0)$ take the form
$$y = mx$$
for some $m \in \Bbb{R}$, when not vertical. When vertical, the equation becomes
$$x = 0.$$
This means two cases. For the first case, substitute $(x, y) = (x, mx)$ into $f(x, y)$:
\begin{align*}
f(x, mx) &= (x^2 + m^2 x^2 - 6x)\cdot\left(\frac{1}{8}m^2x^2 - x\right) \\
&= \frac{1}{8}m^2(m^2 + 1)x^4 - \left(1 + \frac{7}{4}m^2\right)x^3 + 6x^2 
\end{align*}
Let the above function of $x$ be $g(x)$. Now, it's a one variable function, and we can perform the usual tests to check for local minima. For example, we can differentiate:
\begin{align*}
g'(x) &= \frac{1}{2}m^2(m^2 + 1)x^3 - 3\left(1 + \frac{7}{4}m^2\right)x^2 + 12x \\
g''(x) &= \frac{3}{2}m^2(m^2 + 1)x^2 - 6\left(1 + \frac{7}{4}m^2\right)x + 12.
\end{align*}
Note that $g'(0) = 0$, so we have a stationary point, and $g''(0) = 12 > 0$, hence we have a local minimum.
For the second case, we can also consider $f(x, y)$ along the line $x = 0$, i.e. $f(0, y)$, but I'll leave that to you.
However, as mathcounterexamples.net points out, this is not a local minimum.
