Calculate max/min of a 3 variable function, restricted to g(x,y,z)=0 Calculate extrema of $f(x,y,z)=xe^{yz}$ on boundary $3x^2 +y^2 +z^2 =27$


*

*I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + \lambda g_x$,$f_y + \lambda g_y,f_z + \lambda g_z,g(x,y,z)=0$


$e^{yz}+\lambda (6x)=0$
$xze^{yz}+\lambda (2y)=0$
$xye^{yz}+\lambda (2z)=0$
$3x^2 +y^2 +z^2 -27 = 0$
I tried a lot of combinations but I can't solve this. Can you help me ? 

Hint.
Make $\mu = \frac{\lambda}{e^{yz}}$ and then solve
$$
1+\mu 6 x = 0\\
x z + \mu 2 y = 0\\
xy + \mu 2z = 0\\
3x^2+y^2+z^2-27=0
$$
giving
$$
\left[
\begin{array}{ccccc}
x & y & z & \mu & f \\\
 -3 & 0 & 0 & \frac{1}{18} & 0 \\
 3 & 0 & 0 & -\frac{1}{18} & 0 \\
 -\frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & \frac{1}{2 \sqrt{3}} & -\frac{13 e}{\sqrt{3}} \\
 -\frac{1}{\sqrt{3}} & \sqrt{13} & \sqrt{13} & \frac{1}{2 \sqrt{3}} & -\frac{13 e}{\sqrt{3}} \\
 \frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & -\frac{1}{2 \sqrt{3}} & \frac{13 e}{\sqrt{3}} \\
 \frac{1}{\sqrt{3}} & \sqrt{13} & \sqrt{13} & -\frac{1}{2 \sqrt{3}} & \frac{13 e}{\sqrt{3}} \\
\end{array}
\right]
$$
