If \(\mathrm{f}(1)=1, \mathrm{f}^{\prime}(1)=3\), then the derivative of \(\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2\) at \(x=1\) is

\(\begin{aligned} \text { Let } y & =\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2 \\ \therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(x))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(x)) \cdot \mathrm{f}^{\prime}(x)+2 \mathrm{f}(x) \mathrm{f}^{\prime}(x)\end{aligned}\)
\(\begin{aligned}\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=1} & =\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(1))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(1)) \cdot \mathrm{f}^{\prime}(1)+2 \mathrm{f}(1) \mathrm{f}^{\prime}(1) \\ & =3 \cdot 3 \cdot 3+2 \cdot 1 \cdot 3 \\ & =33\end{aligned}\)