Question

Find the equation of the tangent line to the curve
y^3 + yx^2 + x^2 - 3y^2 = 0 at the point (0,3) @Calculus1

Answer 1

differentiating tou get
dy/dx= -(2x-2xy)/(x^2+3Y^2-6y) , substituting (0,3) gives:
dy/dx (slope)=0