$\frac{d y}{d x}+3 \sec ^{2} x \cdot y=\sec ^{2} x$
I.F. $=e^{3 \int \sec ^{2} x d x}=e^{3 \tan x}$
or $y \cdot e^{3 \tan x}=\int \sec ^{2} x \cdot e^{3 \tan x} d x$
or $y \cdot e^{3 \tan x}=\frac{1}{3} e^{3 \tan x}+C$
Given
$y \left(\frac{\pi}{4}\right)=\frac{4}{3}$
$\therefore \quad \frac{4}{3} \cdot e ^{3}=\frac{1}{3} e ^{3}+ C$
$\therefore \quad C = e ^{3}$
Now put $x=-\frac{\pi}{4}$ in equation (1)
$\therefore \quad y \cdot e^{-3}=\frac{1}{3} e^{-3}+e^{3}$
$\therefore y =\frac{1}{3}+ e ^{6}$
$\therefore y \left(-\frac{\pi}{4}\right)=\frac{1}{3}+ e ^{6}$