At what point of the curve

Let the required point be (x1, y1).
The tangent makes an angle of 45o with the x-axis.
∴ Slope of the tangent = tan 45o = 1

Since, the point lies on the curve.

Hence, $y_{1}^{2}=x_{1}$

Now, $y^{2}=x$

$\Rightarrow 2 y \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}$

Given:

$\frac{1}{2 y_{1}}=1$

$\Rightarrow 2 y_{1}=1$

$\Rightarrow y_{1}=\frac{1}{2}$

Now,

$x_{1}=y_{1}^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$

$\therefore\left(x_{1}, y_{1}\right)=\left(\frac{1}{4}, \frac{1}{2}\right)$