Calculus. Show work please

Question

Consider the curve x²y+y²x=6

1. Find dy/dx in terms of x and y.
2. Write the equation for the tangent line where x = 2 and y = 1.
3. Find the coordinates of the point (x, y) where the curve has a horizontal tangent line.

Answer 1

Answer 2

Implicit derivative 2xy+x²dy/dx+y²+2xydy/dx=0 solves to find dy/dx= -y(y+2x)/x(x+2y). The slope at (2,1) = dy/dx = -5/8. Thus the tangent line is y=-5/8x+b. Solve for b at (2,1)=9/4 thus the tangent line is y=-5/8x+9/4.

Horizontal tanget lines are found where dy/dx=0 and solving yields y=0 and y=-2x. Substitute these into the original equatio to find points (0,0) and (3^1/3,-2*3^1/3)