Answer
$$y=\frac{1}{4}x+\frac{3}{4}$$
Work Step by Step
Given $$x+\sqrt{x}=y^{2}+y^{4}, \quad(1,1) $$
Differentiate both sides
\begin{align*}
1+\frac{1}{2\sqrt x} &=2yy'+4y^3y' \\
(2y+4y^3)y'&=1+\frac{1}{2\sqrt x} \\
y'&= \frac{1}{(2y+4y^3)} +\frac{1}{2(2y+4y^3)\sqrt x} \\
\end{align*}
Then
$$m = \frac{ 1}{4} $$
Hence the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-1}{x-1}&= \frac{1}{4} \\
4(y-1)&=(x-1)\\
4y&=x+3\\
y&= \frac{1}{4}x+\frac{3}{4}
\end{align*}
