$\sec (x + y) = xy$

We have, $\sec (x + y) = xy$
On differentiating both sides w.r.t. $x$ , we get
$\frac{d}{{dx}}\sec (x + y) = \frac{d}{{dx}}(xy)$

$\Rightarrow$ $\sec (x + y) \cdot \tan (x + y) \cdot \frac{d}{{dx}}(x + y) = x \cdot \frac{d}{{dx}}y + y \cdot \frac{d}{{dx}}x$

$\Rightarrow$ $\sec (x + y) \cdot \tan (x + y) \cdot \left( {1 + \frac{{dy}}{{dx}}} \right) = x\frac{{dy}}{{dx}} + y$

$\Rightarrow$ $\sec (x + y)\tan (x + y) + \sec (x + y) \cdot \tan (x + y) \cdot \frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y$

$\Rightarrow$ $\frac{{dy}}{{dx}}[\sec (x + y) \cdot \tan (x + y) - x] = y - \sec (x + y) \cdot \tan (x + y)$

therefore,$\frac{{dy}}{{dx}} = \frac{{y - \sec (x + y) \cdot \tan (x + y)}}{{\sec (x + y) \cdot \tan (x + y) - x}}$