Confounding Lagrange multiplier problem 
Optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$

According to the theorem the gradients must be parallell, $\nabla f = \lambda \nabla g$, so their cross product must equal zero i.e. $\nabla f \times \nabla g = \mathbf{0}$. This results in the following system of equations:
$$ \begin{cases} 3y(2y+3x) = 5z(x+2z)  \\ 5z(y + 3z) = 4x(2y + 3x) \\ 4x(x+  2z) = 3y(y+3z)  \end{cases} $$
A system of equations which got me stumped, as I am unable to derive anything useful/sensible from it. Yes I did attempt to incorporate the constraint, albeit unsuccessfully.Now since Lagrange multipliers are failing me (?), I have no clue as to how I can progress. Any help?

Let
$$L(x,y,z,\lambda)=f(x,y,z)-\lambda[g(x,y,z)-6]$$
Then your optimization problem is:
$$\left\{\begin{array}{l}0\stackrel{!}{=}\frac{\partial L}{\partial x}=8x-\lambda(y+3z)\\0\stackrel{!}{=}\frac{\partial L}{\partial y}=6y-\lambda(x+2z)\\0\stackrel{!}{=}\frac{\partial L}{\partial z}=10z-\lambda(2y+3x)\\0\stackrel{!}{=}\frac{\partial L}{\partial \lambda}=xy+2yz+3xz-6\end{array}\right.$$
where the first three equations are equivalent to $\nabla f=\lambda\nabla g$, and the last one to the condition $g(x,y,z)=6$. This gives you a much simpler system of equations than the one you proposed above. Can you take it from here?
