Your work is correct, and it seems that your suggested answer is wrong. When the value of the slope is undefined, this points to the possibility that the tangent line is vertical. This is the case for your problem, so the tangent line should be x=2.
To see this more rigorously, you could consider $x$ as a function of $y$ and calculate $\frac{dx}{dy}$ at $(2,-3)$. You should get $0$. Then use this 'gradient' in the equation $m(y-y_1)=x-x_1$.
If the answer you've listed is correct, perhaps you have copied the problem down incorrectly.
It looks like you have to be mores specific about what you mean about when $y=0$
At $(-1,0)$ you have a vertical tangent line $x=-1$ with undefined slope.
At $(0,0)$ you are correct there is not a uniquely defined tangent line. However you can make a uniquely defined tangent line of you parameterize the curve. That is both $x$ and $y$ would be functions of some parameter for example $t$ then the curve will be drawn $(x(t),y(t))$.
In this way of looking at the curve, the self intersection will be two unique $t$-values and you can draw a distinct tangent line for each one and it looks like the slope will actually be defined in that case, but you are adding extra information with the parameterization.
Edit Also I might add that not just any parameterization will work you'd need to pick a differentiable one, that is $\frac{dx}{dt}$ and $\frac{dy}{dt}$ both exist and are not simultaneously zero. This forces no sharp turns along the path you follow. So at the offending point $(0,0)$ you'd need to make sure it follows to the opposite quadrant as $t$ gets larger.
Just to complete what I was saying above I found a differentiable parameterization of the curve
$$
\begin{split}
x&=\tan^2 t -1\\
y&=\tan^3 t- \tan t
\end{split}
$$
Which has derivatives
$$
\begin{split}
\frac{dx}{dt}&=2\tan t\sec^2 t \\
\frac{dy}{dt}&=3\tan^2 t \sec^2 t- \sec^2 t
\end{split}
$$
On the interval $-\frac{\pi}{2} < t < \frac{\pi}{2}$
And the $(0,0)$ point on the graph occurs at the $t$-values $t=-\frac{\pi}{4}$ and $t=\frac{\pi}{4}$
$$
\begin{split}
\frac{dx}{dt}|_{t=\frac{\pi}{4}}&=4 \\
\frac{dy}{dt}|_{t=\frac{\pi}{4}}&=4\\
\frac{dy}{dx}|_{t=\frac{\pi}{4}}&=\frac{dy/dy|_{t=\frac{\pi}{4}}}{dx/dt|_{t=\frac{\pi}{4}}}=1
\end{split}
$$
$$
\begin{split}
\frac{dx}{dt}|_{t=-\frac{\pi}{4}}&=-4 \\
\frac{dy}{dt}|_{t=-\frac{\pi}{4}}&=4\\
\frac{dy}{dx}|_{t=-\frac{\pi}{4}}&=\frac{dy/dy|_{t=-\frac{\pi}{4}}}{dx/dt|_{t=-\frac{\pi}{4}}}=-1
\end{split}
$$
I'm fairly sure this curve is well-studied, but don't know the name of it and just felt like deriving something like this myself :)
Best Answer
The slope of the tangent line at $(1,2)$ being $0$ implies it is the horizontal line $y=2$.
The normal line is perpendicular to the tangent line at $(1,2)$, hence it is the vertical line $x=1$.
Refer to the graph:
$\hspace{1cm}$