Solving equation $y^2-2ln(y)=x^2$

(Assuming you only care about real solutions).

The Lambert ${W}$ function is the inverse to ${xe^{x}}$ (well, that depends on the value of $x$ - you will need to change the branch accordingly. If ${xgeq -1}$, we take the ${W_0}$ branch, otherwise - you take the ${W_{-1}}$ branch. For simplicity, for now I will refer to the whole thing as just ${W}$). That is,

$${W(xe^{x})=x}$$

Notice that your equation

$${y^2 - 2ln(y)= x^2}$$

is equivalent to

$${2ln(y) -y^2 = -x^2}$$

raising $e$ to both sides yields

$${e^{2ln(y) - y^2}=e^{-x^2}=y^2e^{-y^2}}$$

Multiply both sides by ${-1}$ once again to get

$${Rightarrow -y^2e^{-y^2}=-e^{-x^2}}$$

And now use Lambert ${W}$

$${W(-y^2e^{-y^2})=W(-e^{-x^2})=-y^2}$$

And so you get

$${y^2 = -W(-e^{-x^2})}$$

which implies that

$${y = pmsqrt{-W(-e^{-x^2})}}$$

But which branch of ${W}$ do you take? - it turns out in this context you need to take both one at a time, since both will give you valid solutions. Going back to

$${-y^2e^{-y^2}=-e^{-x^2}}$$

A quick cheap way of seeing this is that by looking at the graph of ${-y^2e^{-y^2}}$ you can see it has a range of ${left[-frac{1}{e},0right]}$, and every value in it's range is hit by some $y$ such that ${(-y^2) < -1}$, and also get's hit by some ${y}$ such that ${(-y^2)geq -1}$ also. And so if you take the ${W_0}$ branch - it'll give you the $y$ solution such that ${(-y^2)geq -1}$, and if you take the ${W_{-1}}$ branch it'll give you the solution such that ${(-y^2) < -1}$. Both solutions are important.

In the context of real solutions, notice that the range of ${-y^2e^{-y^2}}$ being ${left[-frac{1}{e},0right]}$ also forces our ${x}$ to be in the domain of ${(-infty,-1]cup [1,infty)}$ (since this domain ensures the range of ${-e^{-x^2}}$ matches accordingly).