The second derivative of a function is not defined on a manifold. There are two ways to get around this, we can move to a chart or we can define a second derivative in terms of vector fields.
Let $p$ be a critical point of $f:M\rightarrow \mathbb R$ and $X$ and $Y$ in $T_xM$. Let $\tilde{Y}$ be a vector field extending $Y$. Then we can define $(d^2f)_p(X,Y):=X\cdot(\tilde{Y}\cdot f)(p)$.
Show that for $f(x,y)=a\sqrt{(1-x^2-y^2)}$ we have $(d^2f)_{(0,0,a)}(x,y)=-a(x^2+y^2)$.
I am a getting a little confused on how to think of $x,y$ as a vector fields, should the vector field not be $ x \frac{\partial}{\partial x}$ and $y \frac{\partial}{\partial y}$? But this gives the wrong answer
$x \frac{\partial}{\partial x}y \frac{\partial}{\partial y}f= x \frac{\partial}{\partial x}\frac{-y}{\sqrt{1-x^2-y^2}}=\frac{x^2y}{(1-x^2-y^2)^{3/2}}$