Equation for the equipotential lines? What is the equation for the equipotential lines in $x$-$y$ plane for a dipole oriented along the $x$ axis?
 A: In three dimensions, the electric potential $V$ of a pure dipole $\mathbf p$ located at the origin is given by
\begin{align}
  V(\mathbf x) = \frac{1}{4\pi\epsilon_0}\frac{\mathbf p\cdot\mathbf x}{|\mathbf x|^3}
\end{align}
If the dipole is oriented along the $x$-axis, then we have $\mathbf p = p\hat{\mathbf x}$ which gives $\mathbf p \cdot\mathbf x = px$.  Moreover, notice that
\begin{align}
  \frac{1}{|\mathbf x|^3} = \frac{1}{(\sqrt{x^2+y^2+z^2})^3} = \frac{1}{(x^2+y^2+z^2)^{3/2}}
\end{align}
Putting this all together gives the following expression for the potential:
\begin{align}
  V(\mathbf x) = \frac{1}{4\pi\epsilon_0}\frac{px}{(\sqrt{x^2+y^2+z^2})^3}
\end{align}
As pointed out in the comments, the equation for an equipotential is then obtained by setting this expression to a constant.  This gives
\begin{align}
  \frac{x}{(\sqrt{x^2+y^2+z^2})^3} = \mathrm{const.}
\end{align}
If one is interested only in the equation for equipotentials in the $x$-$y$ plane, one can set $z=0$ which gives precisely your quoted result.
