Solution: We have \[ A^{-1} + B^{-1} = (A + B) ^{-1} . \] Multiplying the above equation by $A + B$, we get \begin{align*} A ^{-1} + B^{-1} = (A + B) ^{-1} & \implies (A + B) \left( A^{-1} + B^{-1} \right) = I \\ & \implies A A ^{-1} + AB^{-1} + BA^{-1} + BB^{-1} = I \\ & \implies AB^{-1} + BA^{-1} + I = 0 \\ & \implies X + X^{-1} + I = 0 \tag{$\star$} \end{align*} where $X = AB^{-1} $. Multiplying $(\star)$ by $(X - I) X$, we get \begin{align*} & (X - I) X \cdot \left( X + X^{-1} + I \right) = 0 \\ \implies & (X - I) (X^2 + I + X) = 0 \\ \implies & X^3 + X + X^2 - X^2 - I - X = 0 \\ \implies & X^3 - I = 0. \end{align*} Thus, \[ \det X^3 = \det I = 1 \implies \det X = 1. \] Note that we obtained $\det X = 1$ because $A$ and $B$ are real matrices and so is $X$. Thus, \begin{align*} \det (AB ^{-1} ) = 1 \implies \det A = \det B. \end{align*}

Note that we have explicitly used that the matrix should be real. If we take complex matrices $A$ and $B$, then it is not true. For example, take \[ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}, \quad \text{ and }\quad B = \begin{bmatrix} \omega & 0 \\ 0 & \omega \\ \end{bmatrix}, \] where $\omega $ is the cube root of unity, that is, $\omega = \frac{-1 + \sqrt{3} \iota }{2}$. Then, \begin{align*} A^{-1} + B^{-1} & = I + \begin{bmatrix} \omega ^{-1} & 0 \\ 0 & \omega ^{-1} \\ \end{bmatrix} \\ & = I + \begin{bmatrix} \omega ^2 & 0 \\ 0 & \omega ^2 \\ \end{bmatrix} \\ & = \begin{bmatrix} 1 + \omega^2 & 0 \\ 0 & 1 + \omega ^2 \\ \end{bmatrix}. \end{align*} On the other hand, \begin{align*} (A + B) ^{-1} & = \begin{bmatrix} 1 + \omega & 0 \\ 0 & 1 + \omega \\ \end{bmatrix}^{-1} \\ & = \begin{bmatrix} (1 + \omega )^{-1} & 0 \\ 0 & (1 + \omega )^{-1} \\ \end{bmatrix} \\ & = \begin{bmatrix} (1 + \omega^2 ) & 0 \\ 0 & (1 + \omega^2 ) \\ \end{bmatrix}, \end{align*} where the last equality holds because $1 + \omega + \omega ^2 = 0$. More precisely, \begin{align*} (1 + \omega )(1 + \omega^2 ) = 1 + \omega ^2 + \omega + \omega ^3 = 0 + \omega ^3 = 1. \end{align*}