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?
$$
\begin{align*}
(e^{xy})(y)\begin{pmatrix} \frac{dy}{dx}\end{pmatrix} +(e^{xy})(x)+ e^y\begin{pmatrix}\frac{dy}{dx}\end{pmatrix} + ye^y\begin{pmatrix}\frac{dy} {dx}\end{pmatrix}&=0\\
\frac{dy}{dx} &= \frac{-e^{xy}}{ye^{xy} +e^y+ye^y}
\end{align*}
$$

This is an implicit function
$$ e^{xy} \left( x \frac{dy}{dx} + y \right) + e^y \frac{dy}{dx} + ye^y \frac{dy}{dx} = 0 $$
$$\frac{dy}{dx} = \frac{-y e^{xy}}{xe^{xy} + e^y + ye^y} $$ 
