There are infinetely many solutions. For example we can find solutions $f(x)=a+bx^{c}$. It give $a+\frac{b}{c+1}=2a^{2}+\frac{4ab}{4c+1}+\frac{2b^{2}}{8c+1}+\frac{2}{7}$.
Ok! Let $f(x)=x^{3/8}g(x)$, Then 2$\int_{0}^{1}f^{2}(x^{4})dx=\frac{1}{2}\int_{0}^{1}f^{2}(y)y^{3/4}dy=\frac{1}{2}\int_{0}^{1}g^{2}(y)dy$.
Therefore equation equivalent to \[\int_{0}^{1}(g(x)-x^{3/8})^{2}dx+\frac{4}{7}=\int_{0}^{1}x^{3/4}dx,\]
or \[\int_{0}^{1}(g(x)-x^{3/8})^{2}dx=0.\]
Therefore $g(x)\equiv x^{3/8}$ or $f(x)\equiv x^{3/4}$.