Compute $D_v^2(f)(x_0)$, if $f(x,y)=x^3+x+y^2-2y+1$, $v=\frac{1}{5}(3,4), x_0=(\frac{1}{3},1)$ using the formula $D_v^2(f)(x_0)=v^TH_f(x_0)v$.
So, $v^T= \left[ \begin{array}{cc} \frac{3}{5}\\ \frac{4}{5} \end{array} \right] $ and I found also Hessian matrix to be: $ \left[ \begin{array}{cc} 2&0\\ 0&6 \end{array} \right] $ and $v=(\frac{3}{5}, \frac{4}{5})$ but matrices cannot be multiplied in this way. Would appreciate if somebody could help.
You're almost there, but are misunderstanding the convention. Actually, $$v=\begin{bmatrix} \frac{3}{5} \\ \frac{4}{5}\end{bmatrix}$$ and $$v^T=\begin{bmatrix} \frac{3}{5} & \frac{4}{5}\end{bmatrix}.$$