Find the slope of the given curve at the point $(3,1)$ Find the slope of the given curve at the point $(3,1)$.

$$2y\cos\left(\frac{\pi y}{x}\right)=2x^2-17y$$

How do I start? Differentiate and put the xy values in?
 A: As @AndréNicolas wrote, you will have to use implicit differentiation. Then you can substitute the values for $x$ and $y$ and solve for $y'$. Here is the derivative of the left hand side:
$$\left[2y\cos\left(\frac{\pi y}{x}\right)\right]'$$
$$=[2y]'\cos\left(\frac{\pi y}{x}\right)+2y\left[\cos\left(\frac{\pi y}{x}\right)\right]'$$
$$=2y'\cos\left(\frac{\pi y}{x}\right)+2y\cdot -\sin\left(\frac{\pi y}{x}\right)\cdot \left[\frac{\pi y}{x}\right]'$$
$$=2y'\cos\left(\frac{\pi y}{x}\right)+2y\cdot -\sin\left(\frac{\pi y}{x}\right)\cdot \left(\frac{[\pi y]'\cdot x-\pi y\cdot [x]'}{x^2}\right)$$
$$=2y'\cos\left(\frac{\pi y}{x}\right)+2y\cdot -\sin\left(\frac{\pi y}{x}\right)\cdot \left(\frac{\pi y'x-\pi y}{x^2}\right)$$
You should be able to continue from there.
A: Yes. Evaluating further derivative on both sides
$$y'-y \left(\frac{\pi y'x-\pi y}{x^2}\right) =4x-17y=-5$$
$$y'-  \left({\pi y' 3-\pi}\right)/9 =-5$$
$$y'(1-\pi/3)+(\pi/9 +5) =0$$
$$y'= \dfrac{(\pi/9 +5)}{( \pi/3 -1)}. $$
A: Write the equation as
$$f(x,y)=2y\cos\left(\frac{\pi y}{x}\right)-2x^2+17y=0$$
Using this formula we have
$$\frac{dx}{dy}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$
For the given function we have
$$\frac{\partial f}{\partial x}=\frac{2 \pi  y^2 \sin \left(\frac{\pi  y}{x}\right)}{x^2}-4 x$$
$$\frac{\partial f}{\partial y}=-\frac{2 \pi  y \sin \left(\frac{\pi  y}{x}\right)}{x}+2 \cos \left(\frac{\pi  y}{x}\right)+17$$
so the derivative wrt y is
$$\frac{dx}{dy}=-\frac{\frac{2 \pi  y^2 \sin \left(\frac{\pi  y}{x}\right)}{x^2}-4 x}{-\frac{2 \pi  y \sin \left(\frac{\pi  y}{x}\right)}{x}+2 \cos \left(\frac{\pi  y}{x}\right)+17}$$
Plugging $(3,1)$ we get
$$\frac{36 \sqrt{3}-\pi }{3 \left(18 \sqrt{3}-\pi \right)}$$
