If $x=y+\sqrt{y^2+1}$
Then how to write $y$ in terms of $x$? If $$x=y+\sqrt{y^2+1}$$
Then how to write $y$ in terms of $x$? Please tell me some hints

I assume that 
$x, y \in \Bbb R. \tag 0$
We are given
$x = y + \sqrt{y^2 + 1}. \tag 1$
We observe that this equation implies
$x > 0, \forall y \in \Bbb R, \tag{1.2}$
since
$y^2 < y^2 + 1 \Longrightarrow \vert y \vert < \sqrt{y^2 + 1}; \tag{1.5}$
we proceed:
$x - y = \sqrt{y^2 + 1}; \tag 2$
$x^2 - 2xy + y^2 = (x - y)^2 = y^2 + 1; \tag 3$
$x^2 - 2xy = 1; \tag 4$
$2xy = x^2 - 1; \tag 5$
by virtue of (1.2), we may divide by $2x$:
$y = \dfrac{x^2 - 1}{2x}. \tag 6$
CHECK:
From (6),
$y^2 = \dfrac{(x^2 - 1)^2}{4x^2} = \dfrac{1 - 2x^2 + x^4}{4x^2}; \tag 7$
$y^2 + 1 = \dfrac{1 - 2x^2 + x^4}{4x^2} + 1 = \dfrac{1 - 2x^2 + x^4}{4x^2} + \dfrac{4x^2}{4x^2} = \dfrac{1 + 2x^2 + x^4}{4x^2} = \dfrac{(1 + x^2)^2}{4x^2}; \tag 8$
and again, by virtue of (1.2), we may apply $\sqrt \cdot$ to both sides and find
$\sqrt{y^2 + 1} = \dfrac{1 + x^2}{2x}; \tag 9$
$y + \sqrt{y^2 + 1} = \dfrac{x^2 - 1}{2x} + \dfrac{1 + x^2}{2x} = \dfrac{2x^2}{2x} = x. \tag{10}$
