I have this implicit differentiation problem, taken from James Stewart's Calculus Early Transcendentals, 7th Ed. Page 215, Exercise 15.
Find $\frac{dy}{dx}$ by implicit differentiation. Note that $y$ is a function of $x$.
Given: $$e^\frac{x}{y}=x-y$$
My solution:
Take derivative of left-hand side $$\frac{dy}{dx}e^\frac{x}{y}=(e^\frac{x}{y})\left(\frac{y(1)-(1)y'x}{y^2}\right)=(e^\frac{x}{y})\left(\frac{y-y'x}{y^2}\right)$$
Take derivative of right-hand side
$$\frac{dy}{dx}(x)-\frac{dy}{dx}(y)=1-1(y')=1-y'$$
The expression is now
$$(e^\frac{x}{y})\left(\frac{y-y'x}{y^2}\right)=1-y'$$
Multiply both sides by $y^2$
$$e^\frac{x}{y}y-y'x=y^2-y'y^2$$
Rearrange variables
$$ye^\frac{x}{y}-y^2=y'x-y'y^2$$
Factor $y'$
$$ye^\frac{x}{y}-y^2=y'(x-y^2)$$
Factor $(x-y^2)$ out of right-hand side, divide on the left-hand side
$$\frac{ye^\frac{x}{y}-y^2}{(x-y^2)}=y'$$
The correct answer from the textbook is:
$$\frac{y^2-ye^\frac{x}{y}}{y^2-xe^\frac{x}{y}}=y'$$
Where did I go wrong? Thanks for your help.
You have a mistake in the step "Multiply both sides by $y^2$". Note that
$$y^2\cdot e^{x/y}\frac{y-xy'}{y^2}=e^{x/y}y-e^{x/y}xy'$$ while you have written
$$e^\frac{x}{y}y-y'x.$$