$$\frac { d } { d x } [ x y ] + \frac { d [ \tan x ] } { d x } = \frac { d [ y ^ { 2 } ] } { d x }$$
$\left\{\begin{matrix}x=-\arccos(\sqrt{-\frac{1}{y}})+2\pi n_{2}+\pi \text{, }n_{2}\in \mathrm{Z}\text{, }\exists n_{5}\in \mathrm{Z}\text{ : }\left(n_{2}\geq \frac{n_{5}}{2}-\frac{1}{4}\text{ and }n_{2}<\frac{n_{5}}{2}+\frac{1}{4}\right)\text{; }x=\arccos(\sqrt{-\frac{1}{y}})+2\pi n_{1}-\pi \text{, }n_{1}\in \mathrm{Z}\text{, }\exists n_{5}\in \mathrm{Z}\text{ : }\left(n_{1}>\frac{n_{5}}{2}+\frac{3}{4}\text{ and }n_{1}\leq \frac{n_{5}}{2}+\frac{5}{4}\right)\text{, }&-\sqrt{-\frac{1}{y}}\geq -1\text{ and }y\leq -1\\x=\arccos(\sqrt{-\frac{1}{y}})+2\pi n_{4}\text{, }n_{4}\in \mathrm{Z}\text{, }\exists n_{5}\in \mathrm{Z}\text{ : }\left(n_{4}>\frac{n_{5}}{2}+\frac{1}{4}\text{ and }n_{4}\leq \frac{n_{5}}{2}+\frac{3}{4}\right)\text{; }x=-\arccos(\sqrt{-\frac{1}{y}})+2\pi n_{3}\text{, }n_{3}\in \mathrm{Z}\text{, }\exists n_{5}\in \mathrm{Z}\text{ : }\left(n_{3}\geq \frac{n_{5}}{2}+\frac{1}{4}\text{ and }n_{3}<\frac{n_{5}}{2}+\frac{3}{4}\right)\text{, }&y\leq -1\text{ and }\sqrt{-\frac{1}{y}}\leq 1\end{matrix}\right.$
$y=-\frac{1}{\left(\cos(x)\right)^{2}}$
$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}$