5. [5] The function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f\left(x^{2}\right) f^{\prime \prime}(x)=f^{\prime}(x) f^{\prime}\left(x^{2}\right)$ for all real $x$. Given that $f(1)=1$ and $f^{\prime \prime \prime}(1)=8$, determine $f^{\prime}(1)+f^{\prime \prime}(1)$.
Answer: 6. Let $f^{\prime}(1)=a$ and $f^{\prime \prime}(1)=b$. Then setting $x=1$ in the given equation, $b=a^{2}$. Differentiating the given yields
$$
2 x f^{\prime}\left(x^{2}\right) f^{\prime \prime}(x)+f\left(x^{2}\right) f^{\prime \prime \prime}(x)=f^{\prime \prime}(x) f^{\prime}\left(x^{2}\right)+2 x f^{\prime}(x) f^{\prime \prime}\left(x^{2}\right) .
$$

Plugging $x=1$ into this equation gives $2 a b+8=a b+2 a b$, or $a b=8$. Then because $a$ and $b$ are real, we obtain the solution $(a, b)=(2,4)$.