Answer
We prove that
$f\left( {x,y} \right) = \left( {x + 2y} \right){{\rm{e}}^{xy}}$ has no critical points.
Work Step by Step
We have $f\left( {x,y} \right) = \left( {x + 2y} \right){{\rm{e}}^{xy}}$.
The partial derivatives are
${f_x} = {{\rm{e}}^{xy}} + y\left( {x + 2y} \right){{\rm{e}}^{xy}}$
$ = \left( {1 + xy + 2{y^2}} \right){{\rm{e}}^{xy}}$
${f_y} = 2{{\rm{e}}^{xy}} + x\left( {x + 2y} \right){{\rm{e}}^{xy}}$
$ = \left( {2 + 2xy + {x^2}} \right){{\rm{e}}^{xy}}$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = \left( {1 + xy + 2{y^2}} \right){{\rm{e}}^{xy}} = 0$
${f_y} = \left( {2 + 2xy + {x^2}} \right){{\rm{e}}^{xy}} = 0$
Since ${{\rm{e}}^{xy}} \ne 0$, we have
(1) ${\ \ \ \ }$ $1 + xy + 2{y^2} = 0$
(2) ${\ \ \ \ }$ $2 + 2xy + {x^2} = 0$
From equation (1), we obtain
$xy = - 1 - 2{y^2}$
Substituting $xy = - 1 - 2{y^2}$ in equation (2) gives
$2 + 2\left( { - 1 - 2{y^2}} \right) + {x^2} = 0$
$2 - 2 - 4{y^2} + {x^2} = 0$
$4{y^2} = {x^2}$
$y = \pm \frac{1}{2}x$
Substituting $y = \frac{1}{2}x$ in equation (1) gives
$1 + x\left( {\frac{1}{2}x} \right) + 2{\left( {\frac{1}{2}x} \right)^2} = 0$
$1 + \frac{1}{2}{x^2} + \frac{1}{2}{x^2} = 0$
${x^2} = - 1$
Since $y = \frac{1}{2}x$, so ${y^2} = \frac{1}{4}{x^2} = - \frac{1}{4}$.
However, there are no real solutions for ${x^2} = - 1$ and ${y^2} = - \frac{1}{4}$.
Substituting $y = - \frac{1}{2}x$ in equation (1) gives
$1 + x\left( { - \frac{1}{2}x} \right) + 2{\left( { - \frac{1}{2}x} \right)^2} = 0$
$1 - \frac{1}{2}{x^2} + \frac{1}{2}{x^2} = 0$
$1 = 0$
Therefore, we conclude that $f\left( {x,y} \right) = \left( {x + 2y} \right){{\rm{e}}^{xy}}$ has no critical points.
