Vtyjkyuk

Just wanted to check if this was right before I proceed

f(x,y)=$2x^3 + 6xy^2 - 3y^3 - 150x$

which gives

$frac∂f∂x$ = $6x^2 + 6y^2 -150$

Then doing the same with y gives

$frac∂f∂y$ = $12xy -9y^2$

To find the stationary points, I have to make the derivatives 0 which gives me

$frac∂f∂x$ = $6x^2 + 6y^2 -150$ = $0$
and $frac∂f∂y$ = $12xy -9y^2$ = $0$

rearranging $frac∂f∂x$= $0$ gives me $y^2 = -x^2 + 25$

I proceed to sub this into $frac∂f∂y$ = $12xy -9y^2$ = $0$

This gives me $-12x^2 + 60x + 9x^2 - 225 = 0$

and putting it into quadratic eqtn and then factorising, I get $x=5$ and $x=15$

This is where I get confused, do these numbers sound right and if so, do I place the x coordinates into the original eqtn to get y coordinates?

how many stationary points in total?

You are substituting incorrectly. Here is the right way for $f(x,y)=2x^3 + 6xy^2 - 3y^3 - 150x$:
$$begincasesf_x=6x^2+6y^2-150=0\
f_y=12xy-9y^2=0endcases Rightarrow begincasesx^2+y^2=25\ y(4x-3y)=0endcases Rightarrow \
1) y=0 Rightarrow x^2+0^2=25 Rightarrow x=pm5;\
2) 4x-3y=0 Rightarrow x=frac34y Rightarrow left(frac34yright)^2+y^2=25 Rightarrow y^2=16 Rightarrow y=pm 4 Rightarrow x=pm3.$$
Hence, the stationary points are:
$$(x,y)=(5,0), (-5,0), (3,4), (-3,-4).$$
Note that once you set first order derivatives equal to zero, you must solve the system of equations to find $(x,y)$.

Your approach for finding the stationary points is right unless you also have to check the hessian matrix in the points obtained. For example let $$g(x,y)=x^2-y^2$$the only candidate for stationary point is when $(x,y)=(0,0)$ but we know that this point is not stationary (it is a saddle point). For the candidate you obtained we have $$H=beginbmatrixdfracpartial^2fpartial x^2&dfracpartial^2fpartial xpartial y\dfracpartial^2fpartial xpartial y&dfracpartial^2fpartial y^2endbmatrix_(x,y)=(5,15)=beginbmatrix12x&12y\12y&12x-18yendbmatrix$$if a point is supposed to be stationary we must have the hessian matrix semi positive definite in that point. Surely $x=15$ is invalid since $y^2$ becomes negative therefore we have two candidates$$(5,0),(-5,0)$$and $$H_(5,0)=60I>0\H_(-5,0)=-60I<0$$so $(5,0)$ is the only stationary point.

Here is a depict of the function