Doubt in a property of Laplace equation One of the Laplace equation's property says that the maxima and minima can only occur at the boundaries.
Okay so lets take 2 positive charges, one at the origin and the other $d$ distance apart on the $x$-axis.
So the potential between them would be somewhat like what I have drawn in the image.

Now lets take a region between
$x=d/3$ and $x=2d/3$. Now apply the Laplace equation here in the region, (as there is no charge in this region), and so the potential's maxima and minima should occur at the boundary. But its maxima occurs at $x=d/2$ ??

(Assuming that you're referring to the $3$D case and not the $1$D case).
In your example, $V(r)$ is not an extremum. Let's place two unit charges at $x = \pm d/2$, and look at the resulting potential $V(x,y)$ (taking $V(r \to \infty) = 0$).
At $x = 0$, the potential is $V(0,0) = \frac{2}{d} + \frac{2}{d} = \frac{4}{d}$
Along the $x$ axis (close to $x = 0$), we have
\begin{align}
V(x=\varepsilon, 0) &= \frac{1}{\frac{d}{2} + \varepsilon} + \frac{1}{\frac{d}{2} - \varepsilon}\\ &= \frac{2}{d} \left(\frac{1}{1 + \frac{2 \varepsilon}{d}}  + \frac{1}{1 - \frac{2 \varepsilon}{d}}\right)\\
&= \frac{4}{d} \left(1 + 4 \frac{\varepsilon^2}{d^2} + o\left(\frac{\varepsilon^2}{d^2}\right) \right) > V(0, 0).
\end{align}
But along the y axis, we have:
\begin{align}
V(x=0, y=\varepsilon) &= \frac{1}{\sqrt{\frac{d^2}{4} + \varepsilon^2}} + \frac{1}{\sqrt{\frac{d^2}{4} + \varepsilon^2}}\\ &= \frac{4}{d} \frac{1}{\sqrt{1 + \frac{4 \varepsilon^2}{d^2}}}\\
&= \frac{4}{d} \left(1 - 2 \frac{\varepsilon^2}{d^2} + o\left(\frac{\varepsilon}{d}^2\right) \right) < V(0, 0).
\end{align}
So while the partial derivative along $x$ and $y$ is indeed zero, this corresponds neither to a maximum nor a minimum of $V$ but rather to a saddle point.
