6. a) The equation of a curve is defined implicitly is x^3y^2 = -3xy. Verify that the point (-1,-3) belongs to the curve. Find an equation of the tangent line to the curve at this point.

My work:

Step 1. Verify point. I did and got -9 = -9.

Step 2. Differentiate.

3x^2y^2 + 2x^3yy' = -3x -3xy'

Step 3. Sub pt into above equation.

This is where I am stuck... you see I am able to differentiate it properly but then when it comes time to sub the points in, I get a totally different answer.

I got received tutoring about this question and was able to get it, but a week later here I am reviewing it and cannot. I keep getting dy/dx = 18/3 = 6. It's supposed to be -12 and I got that once, but now I keep getting 6 for some reason. What could I be doing wrong?

My work:

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a) 3x^2y^2dx + 2x^3ydy = -3ydx + -3xdy

b) Put dx's and dy's together. I get (2x^3y+3x)dy = (-3y-3x^2y^2)dx

c) dy/dx = (-3y-3x^2y^2)dx / (2x^3y+3x)dy

d) 9-27 / 6-3 = -18/3 = -6.

e) Teacher solution is -12. How come I get -6?