Lines tangent to graph of equation So I have the following problem and can't seem to arrive at the official answer:

Consider the curve whose equation is $$(2-x)y^2 = x^3$$
  Obtain the equations of the lines tangent to the graph of the curve at the points where $x=3/2$.

My attempt
My first thought is that you can isolate y in this equation and then have $y$ as a function of $x$ by the function $$f(x) = \sqrt{(x^3/(-x+2)^2)}$$
But when you turn the equation into a function then obviously you can't have a single $x$ with two outputs, therefore you can't have more than one line tangent to the curve at $x=3/2$. I find this confusing. You can have the equation become a function but you can still have the equation with two outputs for one x.
So I decided to just use implicit differentiation here. Then the problem boils down to finding $$y - y(3/2) = y'(3/2)(x - 3/2)$$
So after a fair amount of work you find $y(3/2)$ and $y'(3/2)$. I would share what I got but really I'd like to see if other people come up with other results.
Official answers are $$y = 3\sqrt{3}x - 3\sqrt{3}$$ and $$y = -3\sqrt{3}x + 3\sqrt{3}$$

Note that fo $x\ne 2$
$$(2-x)y^2 = x^3\implies y^2=\frac{x^3}{2-x}\implies 2ydy=-\frac{2(x-3)x^2}{(2-x)^2}dx \implies\frac{dy}{dx}=-\frac{2(x-3)x^2}{2y(2-x)^2}$$
and for $x=\frac32$
$$y^2=\frac{x^3}{2-x}=\frac{\frac{27}{8}}{\frac{1}{2}}=\frac{27}{4}\implies y={\pm} \frac{3\sqrt{3}}2 $$
and we obtain


*

*$(x,y)=\left(\frac32,\frac{3\sqrt{3}}2\right)\implies \frac{dy}{dx}=3\sqrt 3$

*$(x,y)=\left(\frac32,-\frac{3\sqrt{3}}2\right)\implies \frac{dy}{dx}=-3\sqrt 3$
and finally


*

*$\left(y-\frac{3\sqrt{3}}2\right)=3\sqrt 3\left(x-\frac{3}2\right)\implies y=3\sqrt 3x-\frac{9\sqrt{3}}2+\frac{3\sqrt{3}}2\implies y=3\sqrt 3x-3\sqrt{3}$

*$\left(y+\frac{3\sqrt{3}}2\right)=-3\sqrt 3\left(x-\frac{3}2\right)\implies y=-3\sqrt 3x+\frac{9\sqrt{3}}2-\frac{3\sqrt{3}}2\implies y=-3\sqrt 3x+3\sqrt{3}$
