For convenience I assume x-axis as directrix. Variable focus is at $(h,k)$ which should be eliminated by differentiation.
$$ (x-h)^2 + (y-k)^2 =y^2 \tag1 $$
$$ x^2 -2 x h -2 y k + h^2+k^2 =0 \tag2$$
Differentiating with respect to x
$$ y^{\prime } k = x-h \tag3$$
Differentiating with respect to x again
$$ y^{\prime \prime} =1/k\tag4 $$
Solving for $(h,k)$ for elimination
$$(h,k) = ( x- \frac{ y^{\prime }}{y^{\prime \prime }},\,\frac{ {1 }}{y^{\prime \prime }}) \tag5$$
Plug this into (1) and simplify to get the required second order differential equation of arbitrary focus location. The ODE with x-axis as directrix and axis of symmetry not necessarily coinciding with y-axis is given by:
$$ 2 y y^{\prime\prime} = 1+ y^ {{ \prime}^2} \tag6$$
So now recognizing this curvature ratio property ( may be mentioned in old (Chelsea Publishers?) Conics text-books).. The red curve is the parabola evolute.
The radius of curvature of a parabola is double the normal segment length intercepted by the directrix.
$$ \dfrac{ y y^{\prime\prime}} { 1+ y^ {{ \prime}^2}} = \frac{R_d}{R_c}=\frac12\tag7$$
The offset axis case has been numerically computed and verified by CAS It is shown ( BC to arbitrary start point and slope) at right.
So this property can be used beneficially to define parabolic arc differential equation.
Change of axes gives the ODE of parabola whose axis is parallel to x-axis and displaced arbitrarily parallel with respect to $x-$ axis. The focus in general is not on x-axis.
$$ 2 x y^{\prime\prime} = -y^{\prime}(1+y^{{ \prime}^2}) \tag8$$
This is not preferred ODE in the unsymmetric case as there are two distinct values of $y$ for same $x$, it is a mess.
EDIT1:
In hindsight perhaps the easiest way is to get ODEs of all parabolas on same x-axis directrix is Formula17_MW_WA. Curvature ratio
$$ \frac{PD}{PC}= = \frac{R_d}{R_c}=\dfrac{(a t^2+a) \sec\phi }{2a \sec^{3}\phi} =\frac12=\dfrac{ y y^{\prime\prime}} { 1+ y^ {{ \prime}^2}}. \tag9$$
$y'=bx-ay$ is parallel to the directrix and $x'=ax+by$ parallel to the axis. This we can see from comparing to the form I mentioned in the comments. So let these be the new coordinates. The inverse is $x=\frac{ax'+by'}{a^2+b^2}$, $\frac{bx'-ay'}{a^2+b^2}$, transforming your equation into $$x'^2+2g\frac{ax'+by'}{a^2+b^2}+2f\frac{bx'-ay'}{a^2+b^2}+c=0,$$ solving for $y'$ we get $$y'=\frac{a^2+b^2}{2(fa-gb)}x'^2+\frac{ga+fb}{fa-gb}x'+c\frac{a^2+b^2}{2(fa-gb)}=Ax'^2+Bx'+C,$$ making (from the formulas given here) $x'=-\frac{B}{2A}$ or $ax+by=-\frac{ga+fb}{a^2+b^2}$ the axis and $y'=\frac{4AC-B^2-1}{4A}$ or $bx-ay=\frac{c(a^2+b^2)-g^2-f^2}{2(fa-gb)}$ the directrix. The semi latus rectum is $p=\frac1{2A}=\frac{fa-gb}{a^2+b^2}$. For the vertex and focus to undo the transformation we transform the point back. The vertex in the rotated coordinates is $(-\frac{B}{2A},\frac{4AC-B^2}{4A})$ making the vertex in the first coordinates $$(\frac12 \frac{-2 g a^3 f+g^2 a^2 b-2 f^2 b a^2+b c a^4+2 c a^2 b^3+c b^5-f^2 b^3}{(a^2+b^2)^2 (f a-g b)},-\frac12 \frac{-2 g^2 a b^2+f^2 b^2 a-2 f b^3 g+c a^5+2 c a^3 b^2+a c b^4-g^2 a^3}{(a^2+b^2)^2 (f a-g b)})$$ and the focus is $(-\frac{B}{2A},\frac{4AC-B^2+1}{4A})$, using the inverse transformation the focus is $$(\frac12 \frac{b c a^2-2 f a g+g^2 b+c b^3-f^2 b}{(a^2+b^2) (f a-g b)},-\frac12 \frac{c a^3+a f^2+a c b^2-g^2 a-2 f g b}{(a^2+b^2) (f a-g b)}).$$
Best Answer
Your work is correct. If you want you can memorize some formulas:
a parabola of equation $x=ay^2+by+c\;$ has the symmetry axis $s$ parallel to the $x$ axis and its equation is:
$s \quad: \quad y=\dfrac{-b}{2a}=h$
The vertex is at: $ \quad V\quad : \quad \left(x(h),h \right)=(k,h)$
The focus is at : $\quad F \quad :\quad (k+p,h)\; $ with $p=\dfrac{1}{4a}$
and the directrix has equation:
$d \quad : \quad x=k-p$
We can easily see that for your parabola $x=-\frac{1}{4}y^2-y-\frac{1}{2}$ the directrix is the line $x=\frac{3}{2}$.
Note that , as for all the conics , the axis of symmetry is parallel to one of the coordinate axis iff the equation does not contain a mixed term in $xy$.