If

$\int_{0}^{x} f(t) d t=x^{2}+\int_{x}^{1} t^{2} f(t) d t$

$\Rightarrow \quad f(x)=2 x-x^{2} f(x)$

$\Rightarrow \quad f(x)=\frac{2 x}{1+x^{2}}$

$\Rightarrow \quad f^{\prime}(x)=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$

Then,

$f^{\prime}(1 / 2)=\frac{2\left(1-\frac{1}{4}\right)}{\left(1+\frac{1}{4}\right)^{2}}=\frac{3}{2} \times \frac{16}{25}=\frac{24}{25}$