Which integrates to $\displaystyle -\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}+C$. Now you use the initial condition y(0)=1 to find C:
$\displaystyle -\frac{1}{3}e^{-3}=\frac{1}{2}e^{0}+C$
$\displaystyle C=-\frac{1}{3}e^{-3}-\frac{1}{2}$
So your solution is $\displaystyle -\frac{1}{3}e^{-3y}=\frac{1}{2}e^{2x}-\frac{1}{3}e^{-3}-\frac{1}{2}$ which is not one of the options.
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