Answer
$y' = \frac{1}{6xy\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3\tan^2(xy^2 + y) \sec^2(xy^2 + y)} - \frac{3y^2}{6xy + 3} $
Work Step by Step
Differentiate both side with respect to $x$:
$\frac{d}{dx}(\tan^3 (xy^2 + y) ) = \frac{d}{dx}(x)$
Remembering that the notation $\tan^3(x)$ is equivalent to writing $(\tan x)^3$, we can differentiate the right side via two applications of the Chain Rule.
$3\tan^2(xy^2 + y) \times \sec^2(xy^2 + y) \times (x(2yy') + y^2 + y') = 1$
$3\tan^2(xy^2 + y) \sec^2(xy^2 + y) (2xyy' + y^2 + y') = 1$
Distributing on the left hand side gives:
$6xyy'\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3y^2\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3y'\tan^2(xy^2 + y) \sec^2(xy^2 + y) = 1$
Now, we isolate the terms involving $y'$, and solve for $y'$:
$6xyy'\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3y'\tan^2(xy^2 + y) \sec^2(xy^2 + y) = 1 -3y^2\tan^2(xy^2 + y) \sec^2(xy^2 + y)$
$y'\big(6xy\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3\tan^2(xy^2 + y) \sec^2(xy^2 + y) \big) = 1 -3y^2\tan^2(xy^2 + y) \sec^2(xy^2 + y)$
$y ' = \frac{1 -3y^2\tan^2(xy^2 + y) \sec^2(xy^2 + y)}{6xy\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3\tan^2(xy^2 + y) \sec^2(xy^2 + y)}$
$y' = \frac{1}{6xy\tan^2(xy^2 + y) \sec^2(xy^2 + y) + 3\tan^2(xy^2 + y) \sec^2(xy^2 + y)} - \frac{3y^2}{6xy + 3} $
