Implicit differentiation of $e^{x^2+y^2} = xy$ I just want to reconfirm the steps needed to answer this question. Thank you
Find $\dfrac{dy}{dx}$ in the followng:
$$e^{\large x^2 + y^2}= xy$$
I got this so far.

$\newcommand{\dd}{\mathrm{d}}\frac{\dd x}{\dd y} e^{x^2}\cdot e^{y^2} = \frac{\dd x}{\dd y} xy$
$u=e^{x^2}$: $\frac{\dd u}{\dd x}=2x(e^{x^2})$
$v=e^{y^2}$: $\frac{\dd v}{\dd x}=2y(e^{y^2})\frac{\dd y}{\dd x}$
$u=x$: $\frac{\dd u}{\dd x}=1$
$v=y$: $\frac{\dd v}{\dd x}=\frac{\dd y}{\dd x}$
$$\begin{align}
e^{x^2}\cdot2y(e^{y^2})\frac{\dd y}{\dd x} + e^{y^2}\cdot 2x(e^{x^2}) &= x \frac{\dd y}{\dd x} + y\\
e^{x^2}\cdot2y(e^{y^2})\frac{\dd y}{\dd x} - x \frac{\dd y}{\dd x} &= y - e^{y^2}\cdot 2x(e^{x^2})\\
\frac{\dd y}{\dd x} \left[e^{x^2}\cdot2y(e^{y^2})-x\right] &= y - e^{y^2}\cdot 2x(e^{x^2})\\
\frac{\dd y}{\dd x} &= \frac{y - e^{y^2}\cdot 2x(e^{x^2})}{e^{x^2}\cdot2y(e^{y^2})-x}
\end{align}$$

$$ e^{x^2+y^2} =xy
\\ e^{x^2+y^2}\; [x^2+y^2]' =[x]'\, y+x\,[y]'
\\ e^{x^2+y^2}\; [2x+2y\,\color{blue}{y'}] =[1]\, y+x\,[\color{blue}{y'}]
\\ 2xe^{x^2+y^2}+2ye^{x^2+y^2}\,\color{blue}{y'} =y+x\,\color{blue}{y'}
\\ 2ye^{x^2+y^2}\,\color{blue}{y'}-x\,\color{blue}{y'} =y-2xe^{x^2+y^2}
\\ (2ye^{x^2+y^2}-x)\,\color{blue}{y'} =y-2xe^{x^2+y^2}
\\ \color{blue}{y' = \dfrac{y-2xe^{x^2+y^2}}{2ye^{x^2+y^2}-x}}  $$
