<s>[INST] Find the equation of tangent line to the given curve 
Find the equation of tangent line to $(x^2+y^2)^{3/2} = 2xy$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$

I just can't figure out how to take the derivative of this function!
 [/INST] You can use the implicit function theorem, which states that (in particular)
$$F(x,y)=0, \quad \text{and} \quad F_y(x,y) \neq 0 \quad \Rightarrow \quad \frac{dy}{dx}(x,y)=-\frac{F_x(x,y)}{F_y(x,y)}.$$
First define
$$ F(x,y) \equiv (x^2+y^2)^{3/2}-2xy $$
then calculate
\begin{align}
F_x(x,y) &= \frac{3}{2}(x^2+y^2)^{1/2}2x-2y \\[2ex]
F_y(x,y) &= \frac{3}{2}(x^2+y^2)^{1/2}2y-2x 
\end{align}
and it follows that
$$ \frac{dy}{dx}(x,y)= -\frac{\frac{3}{2}(x^2+y^2)^{1/2}2x-2y}{\frac{3}{2}(x^2+y^2)^{1/2}2y-2x}.$$
Finally, substituting the desired coordinates we get
$$ \frac{dy}{dx}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)= -1.$$
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