I am given the following definition for a multilinear derivative: Let $n,m,\alpha \in \mathbb{N}$. Then the total derivative of order $\alpha$ is given by the map \begin{align} D^\alpha f(y) &: (\mathbb{K}^n)^\alpha \rightarrow \mathbb{K}^m \\ D^\alpha f(y) &(h^1,\dots,h^\alpha)_l = \sum_{j_1,\dots,j_\alpha = 1}^{n}\partial_{j_1}\cdots \partial_{j_\alpha} f_l(y) h_{j_1}^1\cdots h_{j_\alpha}^\alpha \text{ for } l \in \{1,\dots,m\}. \end{align}
Now I am given the function
\begin{align} f(y_1,y_2) \mapsto (y_2,a(1-y_1^2)y_2 - y_1), \end{align} where $a > 0$. Now I am supposed to calculate the following derivate: $f''(f''(f,f),f)$.
First one can see that $f''_1(f,f) =0$. Then I calculated $f''_2(f,f)$ the following way: \begin{align} \partial_1 ^2 f_2 &= - 2ay_2 \\ \partial_1 \partial_2 f_2 &= \partial_2 \partial_1 f_2 = -2ay_1 \\ \partial_2 ^2 f_2 = 0. \end{align} Thus I get \begin{align} f''(y)(h^1,h^2) &= -2ay_2(h^1_1 h^2_1) - 2ay_1(h^1_2 h^2_1 + h_1^1 h_2^2) \\ &= -2a y_2 (h^1_1 h^2_1) - 4ay_1(h_1^1 h_2^2). \end{align} In particular $f''_2 (h,h) = -2a y_2 h_1 h_1 - 4 a y_1 h_1 h_2$, whence \begin{align} f''(f,f) = -2ay_2 y_2^2 - 4ay_1y_2(a(1-y_1^2)y_2 - y_1). \end{align}
But from here I struggle to calculate the desired derivative, so any help is appreciated.