At the point $\left(t,(t+1)^{\frac 32}\right), \ f'(t)= \frac 32 \sqrt{t+1}$. The equation of the tangent to the curve at that point is
$y = \frac 32 \sqrt{t+1}(x - t) + (t+1)^{\frac 32}$.
Now let $x=\frac 43$and $y=3$. Then solve for $t$. You should get two nice integers for answers.
\begin{align}
\frac 32 \sqrt{t+1}\left(\frac 43 - t \right) + (t+1)^{\frac 32} &= 3\\
\sqrt{t+1}(4-3t) + 2(t+1)^{\frac 32} &= 6 \\
(4-3t) + 2(t+1) &= \frac{6}{\sqrt{t+1}} \\
6-t &= \frac{6}{\sqrt{t+1}} \\
etc
\end{align}
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
Equation of tangent line at point $(a,f(a))$ is $y = f(a) + f'(a)(x - a)$, so we have to find $f'(x)$ and than plug in value $a$ into the result.
$f'(x)=(4x^2+6x+7)'=8x+6 \Rightarrow f'(3)=30$
Since $f(a)=f(3)=61$, we may write next tangent line equation:
$y=61+30(x-3) \Rightarrow y=30x-29$
For the second tangent line we have that $f'(x)=\frac{-1}{2\sqrt{10-x}}\Rightarrow f'(6)=\frac{-1}{4}$ ,and $f(6)=2$, so the second tangent line is:
$y=2-\frac{1}{4}(x-6) \Rightarrow y=\frac{-1}{4}x+\frac{7}{2}$