Show that $e^{xy}+y=x-1$ is an implicit solution to the differential equation $\frac{dy}{dx} = \frac{e^{-xy}-y}{e^{-xy}+x}$ I began by using implicit differentiation on $e^{xy}+y=x-1$.
From that I  got: 
$$\left(y+x\frac{dy}{dx}\right)e^{xy}+\frac{dy}{dx}=1$$
Then using algebra I got to the point where I had this equation:
$$\frac{dy}{dx}=\frac{1-ye^{xy}}{xe^{xy}+1}$$
I'm not sure if I messed up somewhere along the road, or if my final equation is actually the equivalent to $\dfrac{dy}{dx} = \dfrac{e^{-xy}-y}{e^{-xy}+x}$ but I would like help knowing either where I went wrong or how to convert my equation to the final answer.

we have $$e^{xy}y+e^{xy}y'x+y'=1$$so we get
$$y'(e^{xy}x+1)=1-ye^{x}$$
$$y'=\frac{1-ye^{xy}}{1+x^{xy}}$$
multiplying numerator and denominator by $$e^{-xy}$$
we get
$$y'=\frac{e^{-xy}-y}{e^{-xy}+x}$$
this is what we want to prove
