$f"\left(x\right)=\lambda \left(x-1\right)$

Integrating both the side with respect to $x$

$\Rightarrow f^{\text{'}}\left(x\right)=\lambda \frac{{\left(x-1\right)}^2}{2}+c$

As $f^{\text{'}}\left(1\right)=0\Rightarrow c=-2\lambda$
$f^{\text{'}}\left(x\right)=\lambda \frac{{\left(x-1\right)}^2}{2}-2\lambda$

Integrating both the side with respect to $x$

$f\left(x\right)=\frac{\lambda {\left(x-1\right)}^3}{6}-2\lambda x+u$

$f\left(1\right)=-6⟹-2\lambda +u=-6$
$f\left(-1\right)=10⟹\frac{2\lambda }{3}+u=10$
$\therefore \lambda =6$ and $u=6$
$f^{\text{'}}\left(x\right)=0⟹x-1$ or $x=3$

$f\text{'}\text{'}\left(3\right)>0$
$\therefore x=3$, is a point of local minima.