I'm minimizing a function in 2 variables and I have to find the stationary points, so I put both first derivatives equal to 0, and I obtain this:
$$ \begin{eqnarray} {2e^{(2x+y)}+4x-y-3=0} \\ {e^{(2 x + y)} - x + 2 y=0} \end{eqnarray} $$
And, according to all my math knowledge, in $\mathbb{R}^n$ it has no solution. How should I go on from here?
Well, from the second equation you get $e^{2x+y} = x-2y$ so, going back to the first equation, you get $2x-4y+4x-y-3 = 0$, i.e. $y = \frac 65 x -\frac35$. So it amounts to solve $$ e^{\frac{16}{5}x-\frac 35}-x +\frac{12}{5}x-\frac 65 = 0 $$
which has in fact one solution... $x \approx 0.173704$. So, there is a single stationary point $(x^*,y^*)\approx (0.173704 , -0.391555)$. Assuming that the equations you wrote down are exactly the partial derivatives and not some simplification, you will see that the Hessian matrix is positive definite at the stationary point and so the stationary point is a local minimum. In fact you can see that the Hessian matrix is positive definite in $\mathbb{R}^2$ and so the function in strictly convex and the stationary point is a global minimum.