Use implicit differentiation. The RHS will require the product rule.
$\displaystyle \frac{d}{dx}(2x^3+2y^3) = \frac{d}{dx}(9xy)$
Looking at the RHS:
$\displaystyle u = 9x \: \rightarrow \: u' = 9$
$\displaystyle v = y \: \rightarrow \: v' = \frac{dy}{dx}$
$\displaystyle \frac{d}{dx}(9xy) = 9y + 9x \frac{dy}{dx}$
The LHS is simpler:
$\displaystyle \frac{d}{dx}(2x^3+2y^3) = 6x^2 + 6y^2 \frac{dy}{dx}$
Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent