= Solution
The initial point has $H=1/4$. On the branch leaving $(1/2,0)$ toward negative $y$,
$$
x^2=\frac14-y^2+y^4=\left(\frac12-y^2\right)^2,
\qquad x=\frac12-y^2.
$$
Since $\dot y=-x$, the required scalar equation is
$$
\dot y=y^2-\frac12,
\qquad y(0)=0.
$$
Its solution is
$$
\boxed{y(t)=-\frac1{\sqrt2}\tanh\left(\frac{t}{\sqrt2}\right)}.
$$
At $(x,y)=(1/4,-1/2)$, this requires $\tanh(t/\sqrt2)=1/\sqrt2$. Using the <inverse hyperbolic tangent>,
$$
\boxed{t=\sqrt2\,\operatorname{artanh}\frac1{\sqrt2}
=\sqrt2\log(1+\sqrt2)}.
$$
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