Is $\left|\frac{d\vec{r}}{dt}\right| = \frac{d|\vec{r}|}{dt}\;$? Is magnitude of instantaneous velocity same as instantaneous speed? 
More specifically, is $$\left|\frac{d\vec{r}}{dt}\right| = \frac{d|\vec{r}|}{dt}\; $$
Also Is it wrong to say that $\dfrac{d|\vec{v}|}{dt}$ is rate of change of speed?
 A: Your equation is not valid, see Figure below 
With equations  
\begin{equation}
\Vert\mathbf{r}\Vert^{2}=\mathbf{r}\circ \mathbf{r} \Longrightarrow  2\cdot \Vert\mathbf{r}\Vert\cdot d\Vert\mathbf{r}\Vert =2\cdot\left(\mathbf{r}\circ d\mathbf{r}\right)\Longrightarrow d\Vert \mathbf{r}\Vert =\dfrac{\mathbf{r}}{\Vert\mathbf{r}\Vert}\circ d \mathbf{r}
\tag{01} 
\end{equation}
so
\begin{equation}
\bigl|d\Vert\mathbf{r}\Vert\bigr| \le \dfrac{\Vert\mathbf{r}\Vert}{\Vert\mathbf{r}\Vert}\cdot\Vert d \mathbf{r}\Vert =\Vert d \mathbf{r}\Vert
\tag{02} 
\end{equation}
 
Figure in "CLASSICAL MECHANICS" Herbert Goldstein-Charles Poole-John Safko, 3rd Edition 2000, Addison-Wesley (user12262) 
A: Actually you're asking two different questions.


*

*Is the magnitude of instantaneous velocity the same as instantaneous speed? Well, yes, that's the definition of instantaneous speed.

*Is this equation true?
$$\biggl\lvert\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\biggr\rvert = \frac{\mathrm{d}\lvert\vec{r}\rvert}{\mathrm{d}t}$$
No, it's not - but instantaneous speed is the quantity on the left. The one on the right is the radial component of velocity in a circular coordinate system, and it is useful for some detailed calculations, but it's not one of the "basic" kinematic quantities (for most reasonable definitions of "basic").
For fun: an example that shows the difference is uniform circular motion, where the quantity on the right is zero but the one on the left is not. Also note that the thing on the right can actually be negative, if the particle is getting closer to the origin over time.
Since $\vec{v} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ by definition, the quantity on the left in the above equation is $\lvert\vec{v}\rvert$, and so $\frac{\mathrm{d}\lvert\vec{v}\rvert}{\mathrm{d}t}$ is the rate of change of speed.
A: Obviously not: think to a very simple (2d) example, $r(t)=(t,t)$.
Componentwise, the derivative yields $1,1$, and hence $\lvert dr/dt\rvert=\sqrt{2}$.
On the other hand, $\lvert r(t)\rvert = \sqrt{2}\lvert t \rvert$. And the absolute value function is not differentiable in zero.
Hence the two derivative functions coincide almost everywhere, but not in zero where the second does not exist.
A: 
is $\left|\frac{d\vec{r}}{dt}\right| = \frac{d|\vec{r}|}{dt} $ [?]

There's a subtlety here ...
It depends on how (you want that) the right-hand side of your suggested equation is interpreted.
We know, of course, by to the rigorous definition of what the notation you've used is supposed to mean, that the value of a derivative is evaluated "at" a particular value of the (variable) argument; e.g. "at a particular point $t_a$":
$$\left. \frac{d}{dt}[~\vec r] \right|_{t = t_a} := \text{lim}_{\{t \rightarrow t_a\}}\left[~\frac{\vec r[~t~] - \vec r[~t_a~] }{t - t_a}  ~\right].$$
(This, by itself, should be completely unambiguous; and therefore the left-hand side of your suggested equation, too.)
Now, it seems at least formally straight-forward to "take the magnitude" (as in the right-hand side of your suggested equation) and express explicitly:
$$\left. \frac{d}{dt}[~|\vec r|] \right|_{t = t_a} := \text{lim}_{\{t \rightarrow t_a\}}\left[~\frac{|\vec r[~t~]| - |\vec r[~t_a~]| }{t - t_a}  ~\right].$$
But there is a possible ambiguity of interpretation:


*

*Either: all vectors in the above formula are understood in reference to "the limit point" $\vec r[~t_a~]$, at least for the explicit purpose of evaluating the derivative specificly "at the point $t_a$".
In this case, obviously, $|\vec r[~t_a~]| = 0$;
and also (applicable to the left-hand side of your suggested equation) $|\vec r[~t~] - \vec r[~t_a~]| = |\vec r[~t~]|$;
and in consequence your suggested equation holds(1) (separately for each individual "evaluation point $t_a$", and therefore overall, too) for any trajectory $\vec r[~t~]$ (if given by an "intrinsic description").

*Or: all vectors in the above formula are understood in reference a particular "origin" (e.g. $r_C$ as "center of the description"), regardless of any specific "evaluation point $t_a$", but fixed for evaluating the derivatives at any such points.
In this case $\vec r[~t~] \equiv \vec r[~t~] - \vec r_C$;
and $|\vec r[~t_a~] - \vec r_C|$ may or may not be equal to $|\vec r[~t_b~] - \vec r_C|$ for different "evaluation points" $t_a$ and $t_b$;
and in consequence your suggested equation may fail (depending on the detailed trajectory $\vec r[~t~]$ and choice of "origin $r_C$"; with "circular motion in reference to the center" as basic counter-example).
p.s.

[...] speed

Speed is unambiguously the magnitude of velocity; namely expressed by the left-hand side of your suggested equation.
Speed values being therefore necessarily non-negative, one can then distinguish (intrinsically):


*

*the (non-zero) speed in the limit approaching (towards) some particular evaluation point $t_a$, or 

*the (non-zero) speed in the limit departing from some particular evaluation point $t_a$, or

*speed being zero at evaluation point $t_a$.
In reference to some (possibly external) fixed "origin $r_C$" one may also distinguish "radial speed" (corresponding to the suitably interpreted right-hand side of your suggested equation) and "angular speed".
EDIT
(1): At least up to a sign, which would signify a distinction between "speed on the journey before reaching $\vec r[~t_a~]$", and "speed afterwards". However, it seems also questionable whether such a distinction is even intended at all. Accordingly it seems even more appropriate to investigate whether $$ \left\lvert \left\lvert \frac{d}{dt}[~\vec r~] \right\rvert \right\rvert =?\!\!= \frac{d}{d|t|}[~\|\vec r\|~].$$
