Am I doing this one correct?

find dy/dx of $\displaystyle \tan (xy) = y$

$\displaystyle \frac{d}{dx} [y = \tan (xy)]$

$\displaystyle 1\frac{dy}{dx} = \sec^2 (xy)[y+x\frac{dy}{dx}]$

$\displaystyle -x \sec^2 (xy) \frac{dy}{dx} + 1 \frac{dy}{dx} = y\sec^2 (xy)$

$\displaystyle \frac{dy}{dx} = \frac{y\sec^2 (xy)}{1-x\sec^2 (xy)} $