How to show that $$E_r=-\frac{\partial V}{\partial r}$$ and $$E_{\theta}=-\frac1r \frac{\partial V}{\partial \theta}$$
where V is the potential at the point $(r,\theta)$ of the dipole.
I can take the component of the dipole one in the $E_r$ direction and one in $E_{\theta}$ direction and use standard formula but I am unable to get an intution on how the above formula holds.
You could just say that $$ \langle \frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\rangle $$ is what the gradient operator looks like in spherical polar coordinates. Here's an outline of a derivation:
Starting with ${\bf E} = -\nabla V$, imagine an (arbitrary) infinitesimal displacement $d\bf r$. Dot both sides with it to get $$ {\bf E} \cdot d{\bf r} = -(\nabla V) \cdot d{\bf r} = -dV $$ by the definition of gradient. On the other hand, considering $V = V(r, \theta, \phi)$ as a function of position in polar coordinates, $$ dV = \frac{\partial V}{\partial r}dr + \frac{\partial V}{\partial \theta}d\theta + \frac{\partial V}{\partial \phi}d\phi $$ from standard multi-variable calculus. Finally, consider what $d\bf r$ is in polar coordinates: with respect to unit vectors pointing in direction of increasing $r$, $\theta$, and $\phi$, respectively, it is $d{\bf r} = \langle dr, rd\theta, r\sin\theta d\phi \rangle$ (this can be derived from the standard transformations of rectangular to spherical coordinates). Then ${\bf E} \cdot d\bf r$ becomes
$$ {\bf E} \cdot d{\bf r} = E_r dr + E_\theta r d\theta + E_\phi r \sin\theta d\phi $$
Comparing the last two equations, and considering that $d\bf r$ was arbitrary, we must have identical coefficients in front of each coordinates differential:
$$\begin{align} E_r &= -\frac{\partial V}{\partial r} \\ E_\theta &= -\frac{1}{r}\frac{\partial V}{\partial \theta} \\ E_\phi &= -\frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi} \end{align}$$
For a dipole with axis through the poles of the spherical coordinates, the axial symmetry implies that $V$ does not depend on $\phi$ and so $E_\phi = 0$.
EDIT: The following is another point of view, I add it just in case, though I don't think it is what you asked. The potential and field of a dipole can be expressed in purely vector forms that look like this: $$ V = \frac{1}{4\pi\epsilon_0}\frac{\bf p \cdot r}{|r|^3} $$ $$ {\bf E} = -\nabla V = \frac{1}{4\pi\epsilon_0}\left(\frac{3({\bf p\cdot r})\bf r}{|r|^5} - \frac{\bf p}{|r|^3}\right) $$
These are derived using the vector calculus version of Taylor expansion, and give the dominant terms at large distances not just for a dipole consisting of two point charges, but for a generalized dipole - a localized charge distribution with zero net charge and a dipole moment $$ {\bf p} = \int {\bf r} dq = \int {\bf r}\rho({\bf r}) dV $$ where $\rho$ is charge density and the volume integral is over the region of space where the charge is distributed.
Using the above, you could get $E_r$ and $E_\theta$ by dotting $\bf E$ with $\hat r$ and $\hat\theta$, and then convince yourself that the expressions you get are the same as the respective partial derivatives of $V$ when the latter is expressed in terms of $r$ and $\theta$.