When are this two equations equal to zero? Analytical solution I'm trying to get the minimum of a function but I have to find the points where the gradient of this function: $f(x,y)=xy+x \ e^{x+y}$
I need to know when are these 2 equations equal to zero:
\begin{align*}
y + e^{x+y} + x \ e^{x+y} &= 0\\
x + x \ e^{x+y}&=0
\end{align*}
But this may be on my test tomorrow, and I can't apply numerical methods, so I need somewhat analytical solutions. I will be really thankful. Bye!
 A: No point $\{x,y\}\in \mathbb{R}$ exists s.t. it minimizes $f$, which can be seen by looking at the second equation which gives
$$x=\ln\left(-e^{-y} \right)$$ which is complex, when $x\not=0$. So we check for $x=0$, which gives $y=-W(1)\approx-0.567$, where $W(n)$ is the product log function. However, it is easy to check that $f(x,y)$ can attain lower values than for $f(0,-W(1))$, so no global minimum exists. 
A: You would start with the second equation.  If $x\neq 0$ you could divide through by $x$ getting 
$$
e^{x+y}=0$$
which has no (real) solutions.  So for there to be a zero, it must come at $x=0$.
Now plug $x=0$ into the first equation, getting
$$ y = -e^y$$
which has just one real solution, at roughly $(x=0,y=-.5671)$
Without a calculator, you can still tell that there is only one solution, and it comes at $x=0$ and $y$ somewhere between $-0.5$ and $-1$.
EDIT after comment
Yes, I if $x\neq 0$ the situation is even worse, you would need $e^{x+y} = -1$.  THe point that no such solutions exist still is valid.
But the answer given by Lovsovs is better than this answer because he points out that no global minimum exists.  Indeed, at the critical point we have found where the derivatives are zero, the matrix of second derivatives has imaginary eigenvalues, meaning that this point is neither a local minimum nor a local maximum.
