Let y = y ( x ) be the solution of the differential equation...

Given: $\dfrac{dy}{dx} = \dfrac{\tan x + y}{\sin x\left(\sec x - \sin x \tan x\right)}$

$⇒ \dfrac{dy}{dx} = \dfrac{\tan x + y}{\sin x\left(\dfrac{1}{\cos x} - \dfrac{\sin^{2}x}{\cos x}\right)}$

$⇒ \dfrac{dy}{dx} = \dfrac{\tan x + y}{\sin x\left(\dfrac{\cos^{2}x}{\cos x}\right)}$

$⇒ \dfrac{dy}{dx} = \dfrac{\tan x + y}{\sin x \cos x}$

$⇒ \dfrac{dy}{dx} - \dfrac{2y}{\sin 2x} = \sec^{2}x$

$⇒ IF = {e}^{- 2 ∫ cosec2xdx}$

$⇒ IF = {e}^{- \log \left|cosec2x - \cot 2x\right|}$

$⇒ IF = \dfrac{1}{cosec2x - \cot 2x} = cosec2x + \cot 2x$

$⇒ IF = \dfrac{1}{\sin 2x} + \dfrac{\cos 2x}{\sin 2x}$

$⇒ IF = \dfrac{2\cos^{2}x}{2\sin x \cos x}$

$⇒ IF = \cot x$

$⇒ y \times \cot x = ∫ \cot x \times \sec^{2}xdx$

$⇒ y \times \cot x = 2 ∫ cosec2xdx$

$⇒ ycotx = \log \left|cosec2x - \cot 2x\right| + c$

Now, using $y\left(\dfrac{\pi}{4}\right) = 2$ we get,

$⇒ 2cot\dfrac{\pi}{4} = \log \left|cosec\dfrac{\pi}{2} - cot\dfrac{\pi}{2}\right| + c$

$⇒ c = 2$

$⇒ ycotx = \log \left|cosec2x - \cot 2x\right| + 2$

Now, finding $y\left(\dfrac{\pi}{3}\right)$ we get,

$⇒ ycot\dfrac{\pi}{3} = \log \left|cosec\dfrac{2\pi}{3} - cot\dfrac{2\pi}{3}\right| + 2$

$⇒ y \times \dfrac{1}{\sqrt{3}} = \log \left|\dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}}\right| + 2$

$⇒ y \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{2}\log 3 + 2$

$⇒ y = \dfrac{\sqrt{3}}{2}\log 3 + 2\sqrt{3}$

$⇒ y = \sqrt{3}\left(log\sqrt{3} + 2\right)$