#### Answer

The equation of the tangent line is
$$y= x+\frac{1}{2}$$

#### Work Step by Step

Given $$x^{2}+y^{2}=\left(2 x^{2}+2 y^{2}-x\right)^{2}, \ \ \ \ \ (0,\frac{1}{2})$$
Since $x^{2}+y^{2}=\left(2 x^{2}+2 y^{2}-x\right)^{2}$
by differentiate both sides by $x$, we get
\begin{align}
\Rightarrow &2 x+2 y y^{\prime}=2\left(2 x^{2}+2 y^{2}-x\right)\left(4 x+4 y y^{\prime}-1\right)\\
\end{align}
at $(x,y)=(0,\frac{1}{2})$, we get
$0+y^{\prime}=2\left(\frac{1}{2}\right)\left(2 y^{\prime}-1\right) \Rightarrow y^{\prime}=2 y^{\prime}-1 $
so, $$y^{\prime}=1 $$
henc the equation of the tangent line is
$y-\frac{1}{2}=1(x-0)$
i.e
$y= x+\frac{1}{2}$