[tex]\dfrac{\mathrm dy}{\mathrm dx}=(y+2)^2\sin\left(\dfrac xe\right)[/tex]
a. If [tex]y=k[/tex], then [tex]\frac{\mathrm dy}{\mathrm dx}=0[/tex], so
[tex]0=(k+2)^2\sin\left(\dfrac xe\right)\implies (k+2)^2=0\implies k=-2[/tex]
b. When [tex]x=w[/tex], we're told [tex]y[/tex] has a horizontal tangent, which has slope [tex]\frac{\mathrm dy}{\mathrm dx}=0[/tex]. So we have
[tex]0=(y+2)^2\sin\left(\dfrac we\right)\implies\sin\left(\dfrac we\right)=0\implies\dfrac we=2n\pi\implies w=2ne\pi[/tex]
where [tex]n[/tex] is any integer, whose smallest positive value occurs for [tex]n=1[/tex], giving [tex]w=2e\pi[/tex].
c. The equation is separable:
[tex]\dfrac{\mathrm dy}{(y+2)^2}=\sin\left(\dfrac xe\right)\,\mathrm dx[/tex]
Integrate both sides to get
[tex]-\dfrac1{y+2}=-e\cos\left(\dfrac xe\right)+C[/tex]
[tex]y=-1[/tex] when [tex]x=0[/tex], so we find
[tex]-1=-e+C\implies C=e-1[/tex]
Then the particular solution to the DE is
[tex]-\dfrac1{y+2}=-e\cos\left(\dfrac xe\right)+e-1\implies y=\dfrac1{e\left(\cos\left(\frac xe\right)-1\right)+1}-2[/tex]
