How to isolate y in a potential function The following is a potential function:
$Φ(x,y)=x^4y^4+x^2e^xy^2-3lnx-\frac{3}{4}e^2=0$
I would like to isolate $y$ to give $y(x)=…$.
An online calculator gave $y=\pm \frac{\sqrt{-\sqrt{x^4(e^2x+12ln(x)+3e^2}+e^x}x^2}{\sqrt{2}}$, yet I would like to learn how to go about this manually.
Thank you.
 A: As suggested in the comments, the function $\Phi(x,y)$ is quadratic in $y^2$ as
$$\Phi(x,y)=x^4(y^2)^2+x^2e^xy^2-3\ln x-\tfrac{3}{4}e^2=0,$$
and so if $x\neq0$ the quadratic formula gives
\begin{eqnarray*}
y^2
&=&\frac{-x^2e^x\pm\sqrt{(x^2e^x)^2-4x^4(-3\ln x-\tfrac34e^2)}}{2x^4}\\
&=&\frac{-x^2e^x\pm\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}{2x^4}.\\
\end{eqnarray*}
Note that the denominator is positive, so if we have the $-$-sign for the $\pm$-sign then the numerator is negative. But this is impossible because $y^2$ is nonnegative. So we must have the $+$-sign. Then taking square roots shows that
\begin{eqnarray*}
y
&=&\pm\sqrt{\frac{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}{2x^4}}.\\
&=&\pm\frac{\sqrt{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}}{\sqrt{2x^4}}\\
&=&\pm\frac{\sqrt{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}}{x^2\sqrt{2}},
\end{eqnarray*}
which is close to the result you found, but not the same; the denominator is off by a factor $x^2$, and the expression under the outer square root in the numerator is the negative of what you found.
The above can be further simplified to
\begin{eqnarray*}
y
&=&\pm\frac{\sqrt{-x^2e^x+\sqrt{x^4(e^{2x}+12\ln x+3e^2)}}}{x^2\sqrt{2}}\\
&=&\pm\frac{\sqrt{-x^2e^x+x^2\sqrt{e^{2x}+12\ln x+3e^2}}}{x^2\sqrt{2}}\\
&=&\pm\frac{x\sqrt{-e^x+\sqrt{e^{2x}+12\ln x+3e^2}}}{x^2\sqrt{2}}\\
&=&\pm\frac{\sqrt{-e^x+\sqrt{e^{2x}+12\ln x+3e^2}}}{x\sqrt{2}}.
\end{eqnarray*}
