Finding the angle between two tangents of a curve I just got back from my math test. In the test I encountered the following question:

A curve is defined by the equation $$x^3+y^3=3xy$$ Find the angle between the tangents at points (-1,1) and (1,2).

After differentiating implicitly, we get:
$\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$
And the equation of the tangents at the above points are:
$y = \frac{x}{3} + \frac{5}{3}$ and $y = 1$
So, I formed a triangle representing one part of the intersection between the two lines, which looks something like this. 
Therefore, the angle between them would be $$\arcsin(4/5)$$ Is my approach right or wrong? If its wrong then how would we do this question?

Slope formula: $m = \frac{y-x^2}{y^2-x}$
At $P_1 \equiv( -1,1)$, $m_1 = \frac{1-0}{1+1} = 0$
At $P_2\equiv(1,2)$, $m_2 = \frac{2-1}{4-1}=\frac{1}{3}$

Acute angle between the 2 tangents,

$\phi = \tan^{-1}\big\vert\frac{m_1-m_2}{1+m_1m_2}\big\vert = \tan^{-1}\big\vert\frac{0-\frac{1}{3}}{1+0} \big\vert = \tan^{-1}\frac{1}{3}$
