Find the equation of the normal to the curve 2x^3+3xy+2/y=0 at the point (1,-1)

Step 1 Use Implicit differentiation with respect to x and y - 6x^2+3y + 3x(dy/dx) - 2/y^2(dy/dx) = 0Step 2 Write the equation as dy/dx =... - dy/dx = (6x^2 + 3y)/(2/y^2 - 3x)Step 3 Input (1,-1) into the equation to find gradient of tangent. - Gradient of tangent = -3Step 4 Use knowledge the knowledge that the gradient of the normal is equal to the negative reciprocal of the tangent. - Gradient of Normal = 1/3Step 5 Write the equation of the normal in point slop form- y+1 = 1/3(x-1)Step 6 Rearrange so it is in the form y=mx+c - y=1/3(x)-4/3