Find the y-values(s) where $x^2 - y^2 = 3x + 3y + 4$ has vertical tangents 
Find the y-values(s) where $x^2 - y^2 = 3x + 3y + 4$ has vertical tangents

ok, so this is me trying to prove my teacher is wrong....she says that the points are -1,0,1 and I believe this to be incorrect but I'm having issues with deriving the function.
 A: Vertical tangents to a curve occur when 
$$
\frac{dx}{dy}=0
$$
Now for your curve, we implicitly differentiate to find
$$
2x\frac{dx}{dy}-2y=3\frac{dx}{dy}+3\implies \frac{dx}{dy}(2x-3)=3+2y\\
\implies \frac{dx}{dy}=\frac{3+2y}{2x-3}
$$
if this is to vanish, then $3=-2y\implies y=-3/2$.
A: Applying $\frac{d}{dx}$ to the equation, we get:
$$2x-2yy'=3+3y'$$
and then isolating $y'$, we get:
$$y'=\frac{2x-3}{3+2y}.$$
It looks like the derivative should be undefined precisely when $y=-\frac32$
A glance at the graph seems to confirm this.

Another approach would be to complete the squares and see this equation for what it really is:
$$x^2-3x+\frac94-(y^2+3y+\frac94)=4$$
or:
$$\frac{\left(x-\frac32\right)^2}{2^2}-\frac{\left(y+\frac32\right)^2}{2^2}=1.$$
It's a hyperbola with left/right branches, centered at $\left(\frac32,-\frac32\right)$. There's nothing special going on at $y=-1,0,1$.
A: Hint:
The equation of any vertical tangent can be set to $$x=k$$
Now replace this value of $x,$ in the equation of the hyperbola to form a quadratic equation in $y$ whose  roots represent the ordinate of the intersection
Now for tangency, both ordinates should coincide.
Can you take it from here?
