Solving for $x$ in terms of $y$ for equations involving quadratics, logarithms, and exponentials. For the equation:
$$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$
How might one go about solving for $x$ in terms of $y$? First, I attempted solving the equation by first performing the subsitution $u = xe^{xy^2}$, which then reduces the problem to: $$\frac{\ln(\frac{u}{x})}{x} = \frac{\ln(1 - u)}{1-u}$$
Unfortunately, I could not find where to go from there. Secondly, I tried multiplying both sides of the initial equation by $1 - xe^{xy^2}$ to obtain:
$$y^2-xy^2e^{xy^2} = \ln(1 - xe^{xy^2})$$
From here, I could subtract $y^2$ from both sides and divide by -1 to get the equation in a form where the product logarithm could be applied, but unfortunately the right hand side would contain both $x$ and $y$. Exponentiating both sides with base $e$ only seems to make the problem more convoluted and messy than it already is. So how might I go about actually solving this equation for $x$?
 A: Considering $$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$ we could start defining $$z=1 - xe^{xy^2}\implies y^2=\frac{\log(z)}z$$ Solving for $z$
$$z=-\frac{W\left(-y^2\right)}{y^2}$$ where $W(.)$ is Lambert function.
So
$$1 - xe^{xy^2}=-\frac{W\left(-y^2\right)}{y^2}$$ and, again, Lambert function
$$x=\frac{W\left(y^2+W\left(-y^2\right)\right)}{y^2}$$
Do not ask me for $y(x)$, please !
Edit
Since you mention the product logarithm, I suppose that you can run Mathematica. If so,ask for the contour plot of 
$$f(x,y)=y^2 - \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}=0$$ for $-\frac{1}{\sqrt{e}} \leq y \leq \frac{1}{\sqrt{e}}$ and $-1000 \leq x \leq  0$.
A: The substitution:
$$u=x\;\exp \left( x~y^{2}\right) =x\;\exp \left( x~v\right) ;\;v=y^{2}$$ is the the correct Ansatz here. This leads to: 
$$\frac{\log \left( \frac{u}{x}\right) }{x}=\frac{\log \left( 1-u\right) }{1-u}$$ Application of the functional equation on the left hand side of the equation
leads to:
$$\frac{\log \left( u\right) }{x}-\frac{\log \left( x\right) }{x}=\frac{\log
\left( 1-u\right) }{1-u}$$
Now we substitute  $u$ on the left side of the equation:
$$\frac{\log \left( x\;\exp \left( x~v\right) \right) }{x}-\frac{\log \left(
x\right) }{x}=\left( v+\frac{\log \left( x\right) }{x}\right) -\frac{\log
\left( x\right) }{x}=v$$
and recognize, that the term: $\frac{\log \left( x\right) }{x}$ vanishes.
Now, we can solve for $u$
$$u=\frac{v+W\left( -v\right) }{v}$$
where $W\left( z\right) $ is the Lambert function  [Mathematica]
(http://functions.wolfram.com/ElementaryFunctions/ProductLog/27/02/) or
[Corless] (https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf). In the end we
have to substitute  $u\rightarrow x\;\exp \left( x~v\right) $ and $v$ and
solve the equation for $x$
$$x=\frac{W\left( y^{2}+W\left( -y^{2}\right) \right) }{y^{2}}$$ Numerical comparisons for $y=0.316228$ e.g. $v=0.1$ confirms the result.
