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?

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
$$\frac{36 \sqrt{3}-\pi }{3 \left(18 \sqrt{3}-\pi \right)}$$
