Finding $\frac{dy}{dx}$ for $e^{xy} +ye^y=1$ The equation $y'$ for $e^{xy} +ye^y=1$ has $y$ as a function of $x$ which is differentiable. Find $\frac{dy}{dx}$.
My attempt by differentiating w.r.t. $x$ is as follows, could somebody assist to check or point out any mistakes?
$$
\begin{align*}
(e^{xy})(y)\begin{pmatrix} \frac{dy}{dx}\end{pmatrix} +(e^{xy})(x)+ e^y\begin{pmatrix}\frac{dy}{dx}\end{pmatrix} + ye^y\begin{pmatrix}\frac{dy} {dx}\end{pmatrix}&=0\\
\frac{dy}{dx} &= \frac{-e^{xy}}{ye^{xy} +e^y+ye^y}
\end{align*}
$$

When you differentiate both sides with respect to $x$, you should get
$$e^{xy}(y+xy')+y'e^y+y'ye^y=0$$
$$y'=-\frac{ye^{xy}}{e^y+ye^y+xe^{xy}}$$
Just be more careful when you do the calculation.
