Reference: Exercise 1.3.15 of Terry Tao's random matrix theory book at https://terrytao.files.wordpress.com/2011/02/matrix-book.pdf.
Let $A$ be an $n\times n$ Hermitian matrix with $A_{n-1}$ the leading $n-1\times n-1$ submatrix. Let $\lambda$ be an eigenvalue of $A$ that is distinct from all of the eigenvalues of $A_{n-1}$. Suppose that $A_{n-1}=\sum_{j=1}^{n-1}\lambda_{j}v_{j}v_{j}^{*}$ is the eigendecomposition of $A_{n-1}$, and let $X$ denote the vector $(a_{nj})_{j=1}^{n-1}$. Then
$$a_{nn} - \lambda = \sum_{j=1}^{n-1}\frac{|v_{j}^{*}X|^{2}}{\lambda_{j} - \lambda}.$$
The hint is to look at a decomposition of $Au=\lambda u$ into the first $n-1$ components and the last component, but I don't see how to simplify the resulting mess.
Divide $u$ into the vectors $(u_1,z)$ with $u_1 \in \Bbb C^{n-1}$ and $z \in \Bbb C$. We have $$ Au = \lambda u \implies \pmatrix{A_{n-1} & X\\X^* & a_{nn}} \pmatrix{u_1\\z} = \lambda \pmatrix{u_1\\z} \implies\\ A_{n-1}u_1 + zX = \lambda u_1 ; \qquad X^*u_1 + a_{nn}z = \lambda z \implies\\ (a_{nn} - \lambda)z = -X^*u_1; \qquad zX = (\lambda I - A_{n-1})u_1 $$ We ultimately want an expression that makes no reference to $u_1$, so we use the second equation to make the substitution $u_1 = z(\lambda I - A_{n-1})^{-1}X$, noting that the matrix is invertible because $\lambda$ is not an eigenvalue of $A_{n-1}$.
We can now rewrite the first equation as $$ (a_{nn} - \lambda)z = -zX^*(\lambda I - A_{n-1})^{-1}X. $$ Note that $z \neq 0$ since if we had $z = 0$, we would conlcude that $\lambda$ is an eigenvalue of $A_{n-1}$. So, we can divide both sides by $z$. Expanding the right side via $$ (\lambda I - A_{n-1})^{-1} = \sum_{j=1}\frac 1{\lambda - \lambda_j} v_jv_j^* $$ yields the desired expression.