<s>[INST] solve for $y$ in $e^{xy}+x^2+y-1.2$ 
I cant figure out how to make $y$ the subject in this equation: 
  $$e^{xy}+x^2+y-1.2=0$$

I did these steps,
$$\begin{align}&y=1.2-x^2-e^{xy}\\
&\ln(y)=\ln(1.2)-\ln(x^2)-\ln(e^{xy})\\
&\ln(y)=\ln(1.2)-\ln(x^2)-xy\\
&\ln(y)+xy=\ln\frac{1.2}{x^2}\end{align}$$
ahead of this I am lost and would appreciate any and all help.
 [/INST] Let $a=1.2$, rearrange a little and we have
\begin{eqnarray*}
e^{-xy} (a-x^2-y)=1.
\end{eqnarray*}
Multiply by $xe^{ax-x^3}$
\begin{eqnarray*}
 (\color{red}{ax-x^3-xy})e^{\color{red}{ax-x^3-xy}}=xe^{ax-x^3}.
\end{eqnarray*}
Now recall the Lambert $W$ function is defined by $\color{red}we^{\color{red}w}=z$ gives $w=W(z)$. So we have
\begin{eqnarray*}
ax-x^3-xy =W(xe^{ax-x^3}) \\
y=\frac{ax-x^3-W(xe^{ax-x^3})}{x}.
\end{eqnarray*}
</s>