(Too long for a comment.)
No one else has mentioned this, so I will...
Remember: $y$ is a function of $x$, not something that is constant with respect to variation of $x$. (Forgetting this is a fairly common error when first performing implicit differentiation.) It can help to write "$y(x)$" rather than just "$y$".
Your error is believing $\int -x y(x) \, \mathrm{d}x = \frac{-1}{2}x^2 y$. But $y$ is not some constant in this integral; it is a function that depends on $x$. Otherwise, what is written could be the absurdity (when, say $y(x) = \frac{1}{x^2}$), "$\int -x \cdot \frac{1}{x^2} \,\mathrm{d}x = \frac{-1}{2} x^2 \cdot \frac{1}{x^2} + C_1 = C$" rather than the correct
$$ \int -x \cdot \frac{1}{x^2} \,\mathrm{d}x = \ln|x| + C \text{.} $$
Quick way to verify this: Implicitly differentiate your antiderivative to see if you get the integrand and differential element back:\begin{align*}
\left( \frac{-1}{2}x^2 y(x) \right)'
&= \frac{-1}{2}x^2 (y(x))' + \left( \frac{-1}{2}x^2 \right)' y(x) \\
&= \frac{-1}{2}x^2 \,\mathrm{d}y(x) + \left( \frac{-1}{2} \cdot 2 x \,\mathrm{d}x \right) y(x) \\
&= \frac{-1}{2}x^2 \,\mathrm{d}y(x) - x y(x) \,\mathrm{d}x \text{,}
\end{align*}
which isn't quite "$- x y\,\mathrm{d}x$".
(However, we have shown \begin{align*}
\int \left( \frac{-1}{2}x^2 \frac{\mathrm{d}y(x)}{\mathrm{d}x} - x y(x) \right) \,\mathrm{d}x
&= \int \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{-1}{2}x^2 y(x) \right) \, \mathrm{d} x \\
&= \frac{-1}{2}x^2 y(x) + C \text{.}
\end{align*}
Familiarity with this kind of manipulation could be useful in the future.)