On page183 of Rayd'inverno "An introduction to relativity" he says that the right term in parenthesis is a gradient of some scalar field i.e. When $$\partial_a (\frac{\ X_b}{\ X^2})=\partial_b(\frac{ X_a}{\ X^2}) ,$$ $X^2\neq0$ such that $$\frac{X_a}{X^2}=f, _a \tag 1$$ where $f$ is some scalar field.
How equation (1) happened? What is the reason behind it? I don't understand it please somebody explain to me.
OP's eq. (1) follows from Poincare lemma: The co-vector/1-form $$\omega~:=~\frac{ X_a\mathrm{d}x^a}{\ X^2}\tag{A}$$ is closed $$\mathrm{d}\omega~=~0,\tag{B}$$ and hence locally exact, i.e. on the form $$\omega~=~\mathrm{d}f~=~f_{,a}\mathrm{d}x^a\tag{C}$$ of a differential.
Note that a vector $X^a\partial_a$ and a co-vector $X_a\mathrm{d}x^a$ are related via the musical isomorphism. This isomorphism also connects gradient vectors and differentials.
I don't have that book, but if $\partial_a$, $\partial_b$ are Cartesian coordinates, the condition you provided are the coordinates of $\nabla \times \mathbf{F} = \mathbf{0}$,
with $\mathbf{F} = \dfrac{\mathbf{X}}{|\mathbf{X}|^2}$.
It's not hard to prove that $f = \ln \dfrac{|\mathbf{X}|}{X_0} $, by direct computation
$\partial_i f = \partial_i \left( \ln \dfrac{|\mathbf{X}|}{X_0} \right) = \dfrac{X_0}{|\mathbf{X}|} \dfrac{1}{X_0} \partial_i |\mathbf{X}| =\dfrac{1}{|\mathbf{X}|} \partial_i |\mathbf{X}|$
and remembering that $\partial_i |\mathbf{X}|= \frac{X_i}{|\mathbf{X}|}$ we eventually get
$\partial_i \left( \ln \dfrac{|\mathbf{X}|}{X_0} \right) = \frac{X_i}{|\mathbf{X}|^2}$
Given the condition on the curl, you can write $\mathbf{F}$ as the gradient of a scalar field, namely $\mathbf{F} = \nabla f$. Having established that, you can interpret the condition
$\partial_a \left( \dfrac{X_b}{X^2} \right) = \partial_b \left( \dfrac{X_a}{X^2} \right) $$\qquad \rightarrow \qquad$ $\partial_a F_b = \partial_b F_a$$\qquad \rightarrow \qquad$ $\partial_a \partial_b f = \partial_b \partial_a f$
with the Schwartz condition on mixed second derivatives.
