This problem comes from do Carmo's book on page 104. I've almost got this thing worked out but am stuck at one point. The problem is long so I'll try to break it down.
Assume $M$ is a compact Riemannian manifold of even dimension whose sectional curvature is positive.
Show that every killing field $X$ on $M$ has a singularity, i.e. a point $p$ where $X_p = 0$.
We have the following maps:
$A_{X}: \mathfrak{X}(M) \rightarrow \mathfrak{X}(M)$ defined by $A_{X}(Z)=\nabla_{Z}X$
$f: M \rightarrow \mathbb{R}$ defined by $f(q)=\| X \|_{q}^{2}=\langle X, X \rangle_{q}$.
Let $p \in M$ be a critical point of $f$, i.e. where $df_{p}=0$. I proved the following lemma (exercise 2 from p. 104):
Lemma: For any $Z \in \mathfrak{X}(M)$, we have at the critical point $p$
(i) $\langle A_X(Z),X \rangle_p = 0$
(ii) $ \langle A_X(Z),A_X(Z)\rangle_p = \frac{1}{2}Z_p(Z\langle X,X \rangle)+ R(X,Z,X,Z)(p)$
We will look at a point $p \in M$ where $f$ attains its minimum. Assume for contradiction that $X_p \neq 0$. The map $A_X$ defined above induces a map $A:T_pM \rightarrow T_pM$ by $A(y)=A_XY(p)=\nabla_Y X(p)$ where $Y$ is any extension of $y \in T_pM$. Let $E \subset T_pM$ be orthogonal to $X_p$.
I want to show that the restriction $A: E \rightarrow E$ is an antisymmetric isomorphism.
Note it follows from Lemma (i) that $A$ actually gives a map $E \rightarrow E$. I've already showed antisymmetric so I'm going to only show my work on isomorphism. It suffices to show that $A: E \rightarrow E$ is injective.
If $A(y_1)=A(y_2)$ then $A(y_1-y_2)=0$, so let $Z=Y_1 -Y_2$ where $Y_1$, $Y_2$ are extensions of $y_1$ and $y_2$. Then, using Lemma (ii) we have
$0=\langle(A(y_1-y_2),A(y_1-y_2) \rangle_p = \langle A_X(Z),A_X(Z) \rangle_p = \frac{1}{2}Z_p(Z\langle X,X\rangle) + R(X,Z,X,Z)(p)$
By the assumption on curvature, $R(X,Z,X,Z)(p)>0$, so if we can show $Z_p(Z\langle X, X \rangle) \geq 0$ then we will have a contradiction. This is where I'm stuck.
So far you're doing well, but assuming only that $p$ is a critical point. You can assume that $p$ is actually a global minimum, so that the Hessian at $p$ is positive-definite: $$Z_p(Z\langle X,X\rangle) = {\rm Hess}\,f(p)(Z_p,Z_p) \geq 0, $$with equality if and only if $Z_p=0$. Done?