A generalization of L'Hospital's Rule 
Let $\mathcal{O}\subset\mathbb{R}^n$ be open, let $f$, $g:\mathcal{O}\to\mathbb{R}$ be twice continuously differentiable functions, and let $x_0\in\mathcal{O}$. Suppose that $f(x_0)=g(x_0)=0$ and $\nabla f(x_0) = \nabla g(x_0) = 0$. Suppose also that $\nabla^2 f(x_0) = \lambda\nabla^2 g(x_0)$ for some number $\lambda$, and that $\nabla^2 g(x_0)$ is a positive definite Hessian matrix. Prove that
  $$\lim\limits_{x\to x_0} \frac{f(x)}{g(x)} = \lambda$$

Proof
I'm tempted to use the second-order approximation of $f(x)$ and $g(x)$, since that is the only place I can think of that the the hessians might appear. However, the approximation is given by
$$f(x + h) \approx f(x) + \left<\nabla f(x), h\right> + \frac{1}{2}\left< \nabla^2 f(x)h,h\right>$$
I suppose I could say that $x_0 = x + h$, so an approximation for $f(x_0)$ could work, since $f(x_0) = 0$ and $\nabla f(x_0) = 0$. But I doubt that is valid for any $h\neq 0$. I'll see what happens, anyway.
The limit would then become, if I choose $x$ close enough to $x_0$ and $h$ small but nonzero:
$$\lim\limits_{x\to x_0} \frac{f(x)}{g(x)} = \frac{\lim\limits_{x\to x_0} f(x)}{\lim\limits_{x\to x_0} g(x)} = \frac{\lim\limits_{h\to 0} \frac{1}{2} \left< \nabla^2 f(x_0 + h)h, h\right>}{\lim\limits_{h\to 0} \frac{1}{2} \left< \nabla^2 g(x_0 + h)h, h\right>} = \frac{\nabla^2 f(x_0)}{\nabla^2 g(x_0)} = \frac{\lambda \nabla^2 g(x_0)}{\nabla^2 g(x_0)} = \lambda$$
This feels both extremely sketchy and sort of on the right track.
 A: You indeed have the right idea to use a 2nd order Taylor expansion, and work from there. The major thing left is to justify why all the error terms become negligible in the limit so that the quotient is $\lambda$. But before even doing so, you have to verify one small detail; which is that for $h$ sufficiently close, but unequal to $x_0$, we have $g(x_0+h) \neq 0$. This is to ensure that you're not dividing by $0$ anywhere. This is partly where the positive-definiteness assumption on the hessian of $g$ comes into play.
First, let me fix some notation. Write
\begin{align}
f(x_0+h) &= \dfrac{1}{2} D^2f_{x_0}[h,h] + \phi(h)
\end{align}
where $\phi(h)$ is the "remainder" term, and by Taylor's theorem, it satisfies $\lim\limits_{h \to 0}\dfrac{\phi(h)}{\lVert h\rVert^2} = 0$. Note that I simply wrote $D^2f_{x_0}[h,h]$ for what you have written as $\langle \nabla^2f(x_0)h, h \rangle$. Similarly, write
\begin{align}
g(x_0 + h) &= \dfrac{1}{2} D^2g_{x_0}[h,h] + \gamma(h).
\end{align}

Step $1$: Deriving an estimate on $g$.
Note that since $D^2g_{x_0}$ is a positive-definite bilinear form on $\Bbb{R}^n$, this means the number
\begin{align}
\mu := \inf \left\{D^2g_{x_0}[h,h]: \, \, \, h \in \Bbb{R}^n \, \, \, \lVert h\rVert = 1 \right\}
\end{align}
will be strictly positive (by positive-definiteness, each number in the set is positive. By compactness of the sphere, we may use the extreme value theorem to assert that the minimum is actually attained, so $\mu$ is strictly positive). hence, for all $h \in \Bbb{R}^n$, we have that
\begin{align}
D^2g_{x_0}[h,h] = \lVert h\rVert^2 D^2g_{x_0} \left[ \dfrac{h}{\lVert h\rVert}, \dfrac{h}{\lVert h\rVert}\right] & \geq \mu \lVert h \rVert^2 \tag{$*$}
\end{align}
Now, we shall use the fact that $\dfrac{\gamma(h)}{\lVert h\rVert} \to 0$ as $h \to 0$. In terms of $\epsilon$ and $\delta$, we can in particular say that by choosing $\epsilon = \dfrac{\mu}{4} > 0$, there exists a $\delta' > 0$, such that for all $h \in \Bbb{R}^n$, if $0< \lVert h\rVert< \delta'$ then
\begin{align}
\left| \dfrac{\gamma(h)}{\lVert h \rVert^2} \right| < \epsilon = \dfrac{\mu}{4}.
\end{align}
So, by rearranging, we find that
\begin{align}
\left|g(x_0+h) - \dfrac{1}{2}D^2g_{x_0}[h,h] \right| < \dfrac{\mu \lVert h \rVert^2}{4}.
\end{align}
Hence, by the (reverse) triangle inequality and $(*)$, we see that
\begin{align}
|g(x_0+h)| & \geq \left| \dfrac{1}{2} D^2g_{x_0}[h,h] \right| - \dfrac{\mu \lVert h \rVert^2}{4} \\
& \geq \dfrac{\mu \lVert h \rVert^2}{2} - \dfrac{\mu \lVert h \rVert^2}{4} \\
&= \dfrac{\mu \lVert h \rVert^2}{4} \tag{$**$}
\end{align}
Since $\mu > 0$, and $0 < \lVert h \rVert < \delta'$, it follows that this last term is $>0$. This shows that there is a small punctured neighbourhood of $x_0$, such that $g$ is no-where vanishing.

Step $2$: Proving the assertion.
This is a simple matter of manipulating the algebra. Most of the hard-work has been done in step $1$, to prove that everything we are about to do is well-defined. Also, $(**)$ will come in handy shortly. So, for any $h$ satisfying $0 < \lVert h \rVert < \delta'$, we have that
\begin{align}
\left| \dfrac{f(x_0 + h)}{g(x_0+h)} - \lambda \right| &= \left| \dfrac{\frac{1}{2} \lambda D^2g_{x_0}[h,h] + \phi(h)}{\frac{1}{2} D^2g_{x_0}[h,h] + \gamma(h)} - \lambda\right| \\
&= \left| \dfrac{\phi(h) - \lambda \gamma(h)}{\frac{1}{2} D^2g_{x_0}[h,h] + \gamma(h)} \right| \\
& \leq \dfrac{4}{\mu} \left| \dfrac{\phi(h)}{\lVert h \rVert^2} - \dfrac{\lambda \gamma(h)}{\lVert h \rVert^2} \right|, \tag{$\ddot{\smile}$}
\end{align}
where in the final inequality, I made use of $(**)$.
Since $\mu> 0$ is simply a constant, and since the fractions in question approach 0 as $h \to 0$ (Taylor's theorem), it follows that $\left| \dfrac{f(x_0 + h)}{g(x_0+h)} - \lambda\right| \to 0$ as $h \to 0$. This completes the proof. (if you wish, you can transcribe the final part into a fully complete $\epsilon$-$\delta$ argument).

Final Remarks.
If you look at the line above $(\ddot{\smile})$, we see that as $h \to 0$, both the numerator and denominator approach $0$. The purpose of $(\ddot{\smile})$ is to show quantitatively that the numerator goes to $0$ much faster than the denominator. It is precisely this final inequality which shows that the error terms for the quotient is negligible. 
