Equality of Hamiltonian and Lagrangian From the Legendre transform we can deduce that $$\frac{\partial H(q,p)}{\partial q}=-\frac{\partial L(q,\dot{q})}{\partial q}.$$
Similarly can we prove $$\frac{\partial H(q,p)}{\partial p}= \frac{\partial L(q,\dot{q})}{\partial p}~ ?$$
 A: Unless you made a typo, this is not true. In the general case of the Legendre transform: let $F$ be a function of $x$ and $G$ a function of $y$ its Legendre transform defined by:
$$
y = \frac{dF}{dx} \\
G = xy-F
$$
you are asking whether:
$$
\frac{dG}{dy} = \frac{dF}{dy}
$$
You rather have:
$$
\frac{dG}{dy} = x \\
\frac{dF}{dy} = y\frac{dx}{dy}
$$
so you have equality iff $x$ is linear in $y$ i.e. $F$ (or equivalently $G$) is quadratic. While this is false in general, since in a lot of cases you have a quadratic $L$ (or $H$) this is often true (and probably the reason why you thought it was true in the first place perhaps).
Hope this helps.
A: 
Similarly can we prove $$\frac{\partial H(q,p)}{\partial p}= \frac{\partial L(q,\dot{q})}{\partial p}~ ?$$

No, not in general.
But the above equation is true when the velocity is a linear function of the canonical momentum $p$, as shown below.
To see this, you must remember that $L$ is not a normally a function of $p$. This means we can only make sense of the above equation by assuming you have made a substitution for $\dot q$ with a different function, say, $v(q,p)$.
Sometimes it is helpful to use a different letter for the expression for $\dot q$ as a function of $q$ and $p$. So, I write:
$$
\dot q = v(q,p)\;,
$$
where the function $v$ comes from inverting:
$$
\frac{\partial L}{\partial \dot q} = p
$$
For example, for a free particle:
$$
\frac{\partial L_{free}}{\partial \dot q} = m\dot q = p
$$
$$
\to v_{free}(q,p) = p/m
$$
In the above example, $v_{free}$ actually depends only on $p$ and not on $q$. But in general $v$ can depend on both.
Given the above definitions, the Hamiltonian is:
$$
H = pv(q,p) - L(q,v(q,p))
$$
Therefore:
$$
\frac{\partial H}{\partial p} = v(q,p) + p\frac{\partial v}{\partial p} - \frac{\partial L}{\partial \dot q}\frac{\partial v}{\partial q} = v(q,p)
$$
Whereas:
$$
\frac{\partial L(q,v(q,p))}{\partial p} = \frac{\partial L}{\partial \dot q}\frac{\partial v}{\partial p} = p\frac{\partial v}{\partial p}
$$
These are only equal if
$$
v(q,p) = p\frac{\partial v}{\partial p}\;,
$$
which is not true in general.
However, it is true, for example, for a free particle where $v_{free}$ is linearly related to $p$. It's also true for any Lagrangian of the form:
$$
L = \frac{1}{2}m{\dot q}^2 - U(q)
$$
