Show that $f(x,y) = 3xe^y - x^3 - e^{3y}$ has a single critical point and determine whether or not it's an absolute maximum. 
Show that $f(x,y) = 3xe^y - x^3 - e^{3y}$ has a single critical point and determine whether or not it's an absolute maximum.

Solving $f_{x} = 0$ , $f_{y} = 0$ gives $(e^y)^2 = x$ and $(e^y) = x^2$ and $y \neq \ln(x) $ so we presume x = 1. Therefore there's only a single critical point.
It's a local maximum. However, I want to know if it's also an absolute maximum.
So now I have two questions:
a) Is my thought process right so far?
b) From this stage onwards, how do I determine whether this point P(1,0,0) is a local or an absolute maximum?
 A: You're correct that the only critical point is at $x=1$ and it's a local maximum. The question you now need to ask yourself is "Is zero the maximum possible value for this function?". To determine if this is a global maximum, we can look at the derivatives. Particularly, the hessian matrix:
$$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-6x&3e^{y}\\3e^y&3xe^y-9e^{3y}\end{bmatrix}$$
The determinant is then
$$\det(H)=-18x^2e^y+54xe^{3y}-9e^{2y}.$$
For your point to be an absolute maximum, the hessian must be negative everywhere (Do you see why?). For that, we require 
$$18x^2e^y+9e^{2y}>54xe^{2y}$$
Which we can reduce to
$$2x^2+e^y>6xe^y$$
This is not true for all $x,y$, which suggests we might be able to find a point with value greater than $0$. In practice, you could simply try plugging in points. But with a little thought, we can generate such a point. Try to think what values of $x,y$ would make this function larger. Notice that we have $-x^3$ in our function, so large negative values of $x$ might work. To examine the effect of this, let's fix $y=0$. We get
$$f(x,0)=3x-x^3-1$$
Restricting $x<0$, we can write this as
$$f(x,0)=|x|^3-3|x|-1$$
Clearly $f(x,0)\to\infty$ as $x\to-\infty$. So, pick any very negative $x$. Indeed, we see that $f(-10,0)=969$, so your critical point is definitely not an absolute maximum. This approach requires some intuition into the behavior of functions, but it yields a result very quickly.
A: The point $(1,\;0,\;1)$ is a local maximum because 
$f'_x(1,0)=0;\;f'_y(1,0)=0$
$H(x,y)=\left(
\begin{array}{ll}
 -6 x & 3 e^y \\
 3 e^y & 3 e^y x-9 e^{3 y} \\
\end{array}
\right)$
$\det H(x,y)=-18 x^2 e^y+54 x e^{3 y}-9 e^{2 y}$
$\det H(1,0)=27>0$ and $f_{xx}^{(2)}(1,0)=-6<0$
for further details look here.
It is not a global maximum because the function is larger than the tangent plane $z=1$ at point $(1,\;0,\;1)$
Analytically, setting $z=e^y$
$3 x z-x^3-z^3\geq 1$
$x^3-3 x z+z^3+1\leq 0$
RHS can be factored as 
$(z+x+1) \left(z^2 -(x+1)z +x^2-x+1\right)\leq 0$
$z\leq -1-x\to e^y\leq -1-x\land x<-1$
$f(x,y)\geq 1$ for $y\leq \log(-1-x)\land x<-1$
The blue region in the second plot is the area of $xy$ plane where $f(x,y)\geq 1$
Hope this helps
$$...$$


A: we have the system
$$e^{y}=x^2$$ (I) and $$xe^{y}=e^{3y}$$ (II) from (II) we get $$x=e^{2y}$$ and from $$x=(e^{y})^2$$ we get $$x=x^4$$ or $$x(x^3-1)=0$$
