If the solution curve f ( x, y ) = 0 of the differential equation...

The given differential equation is

$\left(1 + \ln \left(x\right)\right)\dfrac{dx}{dy} - xln\left(x\right) = {e}^{y}$

Put $xln\left(x\right) = t$

$⇒ \left(1 + \ln \left(x\right)\right)dx = dt$

$⇒ \dfrac{dt}{dy} - t = {e}^{y}$ which is a linear differential equation.

$I . F = {e}^{- ∫ dy}$

$= {e}^{- y}$

The solution of the differential equation is

$⇒ t \times {e}^{- y} = ∫ {e}^{y} \times {e}^{- y}dy + c$

$⇒ t \times {e}^{- y} = y + c$

$⇒ xln\left(x\right) = y{e}^{y} + c{e}^{y}$

Put $x = 1, y = 0$

$⇒ c = 0$

Put $x = a, y = 2$

$aln\left(a\right) = 2{e}^{2}$

$∴ {a}^{a} = {e}^{2{e}^{2}}$

Hence this is the correct option.