Please refer to my comment to understand why your solution is incorrect.
We have two functions, $f(x)=\frac{1}{2}x^2\implies f'(x)=x$ and $g(x)=\ln x\implies g'(x)=\frac{1}{x}$. Let the $x$-value of the point of tangency on $f(x)$ be $a$ and that of $g(x)$ be $b$.
Let us determine the general equation of a tangent for $f(x)$, then search for specific lines which only touch $g(x)$ once.
\begin{align}
y-y_1&=\dfrac{dy}{dx}(x-x_1)\\
y&=a(x-a)+\frac{1}{2}a^2 \tag{Using the point $(a,f(a)$}\\
\end{align}
Now, we solve for when this tangent intersects a point, $(b,g(b))$ on $g(x)$. Note that this will include solutions where the tangent touches $g(x)$ at more than one point.
\begin{align}
y&=a(x-a)+\frac{1}{2}a^2\\
\ln b&=a(b-a)+\frac{1}{2}a^2\\
\ln b&=ab-a^2+\frac{1}{2}a^2 \\
\ln b&=ab-\frac{1}{2}a^2\tag{1}\\
\end{align}
We must do the same for $g(x)$. We need to find the general equation for the tangent of $g(x)$ and find when they intersect $f(x)$. If you follow the above process, then you should end up with:
\begin{align}
y-y_1&=\dfrac{dy}{dx}(x-x_1)\\
y&=\frac{1}{b}(x-b)+\ln b \tag{Using the point $(b,\ln b$)}\\
\frac{1}{2}a^2&=\frac{1}{b}(a-b)+\ln b \\
\frac{1}{2}a^2 &=\frac{a}{b}-1+\ln b \tag{2}\\
\end{align}
Solving $(1)$ and $(2)$ simultaneously for $a$ and $b$ gives two solutions: $a=0.3982,b=2.511$ and $a=1.774,b=0.5638$, given to four significant figures. Therefore the equations of the tangents are:
$$y=0.3982(x-0.3982)+0.0793$$
and
$$y=1.774(x-1.774)+0.1573$$
Here is a screenshot of Desmos to show the solution.
If the two given conics in $x,y$ are given by the three by three matrices $M$ and $N$, the dual conics are given by the adjugate matrices (which up to a scalar are the inverses $M^{-1}$ and $N^{-1}$ when they are invertible).
The dual conics in $X,Y$ intersect in at most four points $(X_i,Y_i)$. And the correspondence $X_ix+Y_iy+1=0$ give the common tangent lines.
Added:
Duality
In projective geometry there's the concept of duality: two lines have a common point is dual to two points have a common line. This extends to curves in that a line is the points on it has the dual concept that a point has the lines through it.
Now this extends to nonlinear curves by a point on a curve having a tangent picking out one line of the lines through the point. The collection of tangent lines to a curve then is a dual curve.
A common tangent to two curves then is exactly an intersection point of the dual curves.
As it happens, for conics, this is as straightforward as inverting the matrices associated to them. The only complication being that some matrices are not invertible and the process still works with the adjugate matrices then.
Example
The conics in the image are
$$x^2+2xy-y^2+3x+3y+3=0,2x^2+2xy-2y^2-8x+4y+2=0$$
which through $a_ix^2+2h_ixy+b_iy^2+2g_ix+2f_iy+c_i=0$ have matrices
$$\begin{pmatrix} a_i&h_i&g_i\\h_i&b_i&f_i\\g_i&f_i&c_i\end{pmatrix},i=1,2$$
or
$$M=\begin{pmatrix} 1&1&\frac32\\1&-1&\frac32\\\frac32&\frac32&3\end{pmatrix},N=\begin{pmatrix} 2&1&-4\\1&-2&2\\-4&2&2\end{pmatrix}$$
The dual conics are then
$$-\frac{21}{4}x^2-\frac32xy+\frac34y^2+6x-2=0,-8x^2-20xy-12y^2-12x-16y-5=0$$
from the adjugate matrices
$$\begin{pmatrix} +\begin{vmatrix}b_i&f_i\\f_i&c_i\end{vmatrix}&-\begin{vmatrix}h_i&f_i\\g_i&c_i\end{vmatrix}&+\begin{vmatrix}h_i&b_i\\g_i&f_i\end{vmatrix}\\-\begin{vmatrix}h_i&g_i\\f_i&c_i\end{vmatrix}&+\begin{vmatrix}a_i&g_i\\g_i&c_i\end{vmatrix}&-\begin{vmatrix}a_i&h_i\\g_i&f_i\end{vmatrix}\\+\begin{vmatrix}h_i&g_i\\b_i&f_i\end{vmatrix}&-\begin{vmatrix}a_i&g_i\\h_i&f_i\end{vmatrix}&+\begin{vmatrix}a_i&h_i\\h_i&b_i\end{vmatrix}\end{pmatrix}.$$
Now you can get the intersections from wolfram alpha
or Maxima
solve([-21/4 *x^2-3/2 *x*y+3/4 *y^2+6*x-2,-8*x^2-20*x*y-12*y^2-12*x-16*y-5],[x,y]);
[[x = 0.180395566181265, y = - 1.037735255240134],
[x = 0.3091137649277184, y = - 0.6697463463799764],
[x = 1.468166586883676, y = - 1.389356814701378],
[x = 2.787004998077662, y = - 3.732949087415946]]
And the common tangent lines are $$\begin{align}0.180395566181265x-1.037735255240134y+1&=0\\0.3091137649277184x-0.6697463463799764y+1&=0\\1.468166586883676x-1.389356814701378y+1&=0\\2.787004998077662x-3.732949087415946y+1&=0\end{align}$$
Links
See wikipedia for how to put duality in coordinates and then set $z=1$ to get the usual $x,y$-plane.
Also see this which for me was the first google search hit for dual conic.
Best Answer
These ovals have parameterisations
$(x(t),y(t))=(b\frac{ab\cos(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2},a\frac{ab\sin(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2})$ and $(x(t),y(t))=(h+b\frac{ab\cos(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2},k+a\frac{ab\sin(t)}{b^2(\cos(t))^2+a^2(\sin(t))^2}).$
The dual curves are given by $(p(t),q(t))=(\frac{-y'(t)}{x'(t)y(t)-x(t)y'(t)},\frac{x'(t)}{x'(t)y(t)-x(t)y'(t)}).$
Now find the intersection points $P_i$ of the dual curves. Then the common tangents are the lines $x x(P_i)+y y(P_i)=1.$
The picture below shows only a few of the common tangents:
Here are the dual curves. Note that they have nodes corresponding to the bitangents.
To accentuate the node bitangent correspondence, here are both the curve and its dual, the point and the dual line (bitangent).
The red intersection point $P_3: (l,m)$ corresponds to the red common tangent through the correspondence $(l:m:-1)$ to $lx +my-1=0.$
I used $a=0.5, b=2.3, h=2.95, k=0.85$ with the following to get the plot
There are ways to get the implicit equations defining the dual curves (I'd need to read up on them), and you could further use grobner bases.
The first has equation $a^4b^{10}x^6+3a^6b^8x^4y^2+3a^8b^6x^2y^4+a^{10}b^4y^6+(8a^6b^6-8a^4b^8-a^2b^{10})x^4+(-20a^8b^4+38a^6b^6-20a^4b^8)x^2y^2+(-a^{10}b^2-8a^8b^4+8a^6b^6)y^4+(16a^8b^2-32a^6b^4+8a^4b^6+8a^2b^8)x^2+(8a^8b^2+8a^6b^4-32a^4b^6+16a^2b^8)y^2-16a^6b^2+32a^4b^4-16a^2b^6=0$