Here's what I have done:

$\bullet$ Let the Lagrangian $L(q_{i},\dot{q}_{i},t)$, which under the **point transformations**

$$
\{q_{i}\}\leftrightsquigarrow\{Q_{i}\}
$$
given by the invertible relations $Q_{i}=Q_{i}\big(q_{j},t\big)\Leftrightarrow q_{j}=q_{j}\big(Q_{i},t\big)$, $\ \ i,j=1,2,...,n$, (i.e. $\det\Big|\frac{\partial Q_{j}}{\partial q_{i}}\Big|\neq 0$), can be written as:
$$
L(q_{i},\dot{q}_{i},t)\stackrel{}{\leftrightsquigarrow}L(Q_{i},\dot{Q}_{i},t)
$$
$\bullet$ Differentiating the point transformations $Q_{i}=Q_{i}\big(q_{j},t\big)$, we get that:
$$
\frac{dQ_i}{dt}\equiv\dot{Q}_i=\frac{\partial Q_i}{\partial q_j}\dot{q}_i+\frac{\partial Q_i}{\partial t}\ \ \ \ \ \Rightarrow \ \ \ \ \ \frac{\partial\dot{Q}_i}{\partial\dot{q_j}}=\frac{\partial Q_i}{\partial q_j} \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
$$
$\bullet$ On the other hand, differentiating the Lagrangian we get that:
$$
\frac{\partial L}{\partial\dot{q}_i}=\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i} \Rightarrow
$$
$$
\Rightarrow \frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\frac{\partial}{\partial\dot{q}_j}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)= \sum_{m=1}^{n}\frac{\partial}{\partial\dot{Q}_m}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}=
$$
$$
=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}\stackrel{(1)}{\Rightarrow}
$$
$$
\stackrel{(1)}{\Rightarrow}
\frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial Q_k}{\partial q_i}\frac{\partial Q_m}{\partial q_j}
$$
Thus:
$$
\Big[\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big]=\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]^{t}
\Big[\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big]\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]
$$
where $[..]$ stands for the respective Jacobian and hessian matrices and $[..]^t$ stands for the transpo-se matrix. The last relation clearly tells us that the non-vanishing of the hessian determinant is invari-ant under the point transformations, i.e.
$$
\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0 \Leftrightarrow \det\Big|\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big|\neq 0
$$
which concludes the proof.

**P.S.:** We can easily see why the above, is a sufficient condition for a Hamiltonian description to exist: Generalized momenta $p_{i}$, are defined by
\begin{equation} \label{7.55}
p_{i}=\frac{\partial L}{\partial\dot{q}_{i}}=p_{i}\big(q_{j},\dot{q}_{j},t\big), \ \ \ i,j=1,2,...,n
\end{equation}

If the Jacobian of the generalized momenta w.r.t. the generalized velocities, or equivalently the Hessian of the
Lagrangian w.r.t. the generalized velocities
\begin{equation} \label{7.57}
\det\Big|\frac{\partial p_{i}}{\partial\dot{q}_{j}}\Big|=\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}
\partial\dot{q}_{j}}\Big|\neq 0
\end{equation}
is non-zero, then, due to the inverse function theorem in $\mathbb{R}^{n}$, the above relations can be solved w.r.t.
the generalized velocities:
\begin{equation} \label{7.58}
\dot{q}_{j}=\dot{q}_{j}\big(q_{i},p_{i},t\big), \ \ \ i,j=1,2,...,n
\end{equation}
Then, the Hamiltonian function is defined through a **Legendre transform**:
\begin{equation} \label{7.59}
H=H\big(q_{i},p_{i},t\big)=\sum_{i=1}^{n}p_{i}\dot{q}_{i}-L\big(q_{i},\dot{q}_{i},t\big)
\end{equation}
with $i=1,2,...,n$. The Hamiltonian function $H:\mathbb{R}^{2n+1}\rightarrow\mathbb{R}$, is a function,
of generalized coordinates, generalized momenta and time.