<s>[INST] Lagrange Multiplier Three Dimension (2). Consider the surface produced by the equation $xy^2z^3 = 2$ . Find the points in this surface closest to the origin? 
So far here is what I have, 
Let 
$$ f(x,y,z) = x^2 + y^2 + z^2 $$
Where f is the square of the distance from the point $(x,y,z)$ to the origin. We want to minimize $f$, so our constraint is 
$$ g(x,y,z) =  xy^2z^3 - 2 $$
We must solve 
$$\nabla f = \lambda \nabla g \quad \text{ and } \quad  g = 0$$ 
By the former we must have 
$$
\Bigg( \begin{matrix} 2x \\ 2y \\ 2z \\ \end{matrix} \Bigg) = 
\lambda \Bigg( \begin{matrix}
y^2z^3 \\
2xyz^3 \\
3xy^2z^2 \\
\end{matrix} \Bigg) $$
Solving this system we get equations
$$ 2x = \lambda \text{ } y^2z^3  $$ $$ $$ 
$$ 2y = 2\lambda \text{ } xyz^3 $$ $$ $$ 
$$ 2z = 3\lambda  \text{ } xy^2z^2$$ $$ $$

This is where I start to get stuck. From what I gathered from websites and youtube videos is that I would try to solve it using either cases, or by having the left side equal the same thing. I tried the latter and believe the answers were
$$ y^2 = 2x^2 \to y = \pm x\sqrt(2) \text{    and}$$
$$ 2z^2 = 3y^2 \to z = \pm y \sqrt{\frac{3}{2}} $$ 
I got these solutions by multiplying the first, second, and third equation by yz, xz, xy, respectively. Then having the first equation equal the second and then having the second equation = third equation. So it should look like this:
$$ 2xyz_1 = \lambda \text{ } y^3z^4  $$ $$ $$ 
$$ 2xyz_2 = 2\lambda \text{ } x^2yz^4 $$ $$ $$ 
$$ 2xyz_3 = 3\lambda  \text{ } x^2y^3z^2$$ $$ $$
But from here I am stuck and I have no idea what to do or even if I am on the right path. Any help would be greatly appreciated. 
 [/INST] Hint
Note that $xyz\ne 0$ because if one is zero the other two must be zero. Thus we have that
$$\dfrac{2x}{y^2z^3}=\frac{2y}{2xyz^3}=\frac{2z}{3xy^2z^2}.$$ Simplifying the fractions we arrive at
$$\dfrac{2x}{y^2z^3}=\frac{1}{xz^3}=\frac{2}{3xy^2z}.$$ Now we cancel a $z$ in each denominator getting
$$\dfrac{2x}{y^2z^2}=\frac{1}{xz^2}=\frac{2}{3xy^2}.$$ From the first and the third we have $$6x^2y^2=2y^2z^2\implies z^2=3x^2.$$ From the second and the third $$3xy^2=2xz^2\implies 3y^2=2z^2\implies y^2=2x^2.$$ Thus
$$y=\pm\sqrt2 x,\quad z=\pm \sqrt 3 x.$$ Now use the constraint to get 
$$\pm x\cdot 2x^2 \cdot 3\sqrt 3 x^2=2.$$
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