(Q.1)
[tex]y = \dfrac{C - e^x}{2x} \implies y' = \dfrac{-2xe^x-2C+2e^x}{4x^2} = \dfrac{-xe^x-C+e^x}{2x^2}[/tex]
Then substituting into the DE gives
[tex]\dfrac{-xe^x-C+e^x}{2x^2} = -\dfrac{2\left(\dfrac{C-e^x}{2x}\right) + e^x}{2x}[/tex]
[tex]\dfrac{-xe^x-C+e^x}{2x^2} = -\dfrac{C-e^x + xe^x}{2x^2}[/tex]
[tex]\dfrac{-xe^x-C+e^x}{2x^2} = \dfrac{-C+e^x - xe^x}{2x^2}[/tex]
and both sides match, so y is indeed a valid solution.
(Q.2)
[tex]\ln\left(y^x\right)\dfrac{\mathrm dy}{\mathrm dx} = 3x^2y[/tex]
This DE is separable, since you can write [tex]\ln\left(y^x\right)=x\ln(y)[/tex]. So you have
[tex]x\ln(y)\dfrac{\mathrm dy}{\mathrm dx} = 3x^2y[/tex]
[tex]\dfrac{\ln(y)}y\,\mathrm dy = 3x\,\mathrm dx[/tex]
Integrate both sides (on the left, the numerator suggests a substitution):
[tex]\dfrac12 \ln^2(y) = \dfrac32 x^2 + C[/tex]
Given y (2) = e ³, we find
[tex]\dfrac12 \ln^2(e^3) = 6 + C[/tex]
[tex]C = \dfrac12 \times3^2 - 6 = -\dfrac32[/tex]
so that the particular solution is
[tex]\dfrac12 \ln^2(y) = \dfrac32 x^2 - \dfrac32[/tex]
[tex]\ln(y) = \pm\sqrt{3x^2 - 3}[/tex]
[tex]\boxed{y = e^{\pm\sqrt{3x^2-3}}}[/tex]
(Q.3) I believe I've already covered in another question you posted.
