Given the curve $x = \frac{e^y}{3+2y}$ I'm trying to find the equation of the normal to the curve at P($1/3,0$). I'd like to give the answer in the form $y = mx + c $.
I know the normal line has slope $\frac{-1}{y'(x_0)}$, so the equation should be $$y-y_0 = \frac{-1}{y'(x_0)}(x-x_0)$$
So $\frac{dy}{dx} = \frac{1}{dy/dx} = \frac{(3+2y)^2}{e^y(1+2y)}$ but that needs a value of $y$ to be entered, so I think there must be something I'm misunderstanding.
Differentiate implicitly: your function's implicit equation is
$$3x+2xy=e^y\implies3+2y+2xy'=e^yy'\implies(e^y-2x)y'=3+2y\implies$$
$$y'=\frac{3+2y}{e^y-2x}\implies\text{ at the point}\;\left(\frac13,\,0\right)\,,\,\text{ the tangent line's slope is}\;\;\frac{3+0}{1-\frac23}=9$$
so the normal line is $\;y=-\cfrac19\left(x-\cfrac13\right)=-\cfrac19x+\cfrac1{27}\;$