This is an implicit differentiation problem.
2*x*y^2 + 3*LN(y) = x^2  3*y^3
differentiating using the product rule and chain rule for the first half of the left side, and the chain rule for the second half of the left side yields
4xyy' + 2y^2 + (3/y)y'
for the first half of the right side we differentiate as usual and for the last half we use the chain rule, which yields
2x  (9y^2)y'
the whole thing being
4xyy' + 2y^2 + (3/y)y' = 2x  (9y^2)y'
grouping like terms and solving for y' (dy/dx) yields
dy/dx = (2x 2y^2) / (4xy + (3/y) + 9y^2)
Plugging in or point at (3,1) yields
dy/dx = (6  2) / (12 + 3 + 9) = 4/24 = 1/6
Dec 15

Eric O.