Find probability density function of $Y=X^2$ for uniform $X.$ Suppose that a random variable $X$ has a uniform distribution on the interval $[0,1]$. Determine the p.d.f of $X^2$
Following this equation:
$$f(x=y)\frac{dx}{dy}$$
I then solve for $Pr(Y\le y)=Pr(X^2 \le y)=Pr(X\le y^{\frac{1}{2}})$ so that $x=y^{\frac{1}{2}}$
Then taking the derivative of $\sqrt{y}$ I get $\frac{1}{2}y^{-\frac{1}{2}}$
I should get $$y^{\frac{1}{2}}\frac{1}{2}y^{-\frac{1}{2}}=\frac{1}{2}$$
I have taken this from Degroot's Probability and Statistics (third edition- page 164 exercise7.) However the solution is actually
$$1 \cdot \frac{1}{2}y^{-\frac{1}{2}} $$
What is the explanation behind this?
Added comments:
$G(y) = Pr(Y \le y) = Pr[r(X) \le y]=Pr[X \le s(y)] = F[s(y)]$
If it is also assumed that s is a differentiable function over the interval $(\alpha, \beta)$, then the distribution of $Y$ will be continuous and the value of its p.d.f $g(y)$ for $\alpha < y < \beta$ will be:
$g(y) = \frac{dG(y)}{dy}=\frac{dF[s(y)]}{dy}=f[s(y)]\frac{ds(y)}{dy}$
In other words, the pdf $g(y)$ of $Y$ can be obtained directly from the pdf $f(x)$ by replacing $x$ but its expression in terms of $y$ and multiplying the result by $dx/dy$

You seem to be mixing two approaches. Instead you should choose one.
Use the CDF, so a density of $f_x(x)=1$ gives $F_X(x)=P(X \le x)=x$ for $0 \le x \le 1$
and so if $Y=X^2$ and $X=Y^{1/2}$
then $F_Y(y) = P(Y \le y)=P(X^2 \le y)=P(X \le y^{1/2}) = y^{1/2}$ for $0 \le y \le 1$
giving a density of $f_Y(y)=F'_Y(y)=\frac{1}{2y^{1/2}}$  for $0 \le y \le 1$.
Or use a change of variables on the density where $f_Y(y) =f_X\big(g^{-1}(y)\big)  \left| \frac{d}{dy} \big(g^{-1}(y)\big) \right|$
and here $g(x)=x^2$ and $g^{-1}(y)=y^{1/2}$ which is monotonic  for $0 \le y \le 1$,
and $f_X\big(g^{-1}(y)\big)=1$ and $\frac{d}{dy} \big(g^{-1}(y)\big)  =  \frac{1}{2y^{1/2}}$
so $f_Y(y) =1  \left|\frac{1}{2y^{1/2}} \right| = \frac{1}{2y^{1/2}} $ for $0 \le y \le 1$.
These are the same density, as you might hope.
