If $\boldsymbol{A}$ is a symmetric second-order tensor, its eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$ are all real. The eigenvalues are related to the invariants $I_1$, $I_2$, $I_3$ of $\boldsymbol{A}$ through the characteristic polynomial
$$
-\lambda^3 + I_1 \lambda^2 - I_2 \lambda + I_3 = 0 \, ,
$$
of which they are the (real) roots.
The roots of this polynomial can be obtained using Cardano's method. With the change of variable $\lambda = X + \frac{1}{3}I_1$, the previous equation rewrites as
$$
X^3 + p X + q = 0
$$
$$
\text{with}\quad p=I_2 - \frac{{I_1}^2}{3}, \quad q= \frac{I_1 I_2}{3} - \frac{2 {I_1}^3}{27} - I_3.
$$
All eigenvalues are real if the discriminant $\Delta = - (4 p^3 + 27 q^2)$ is positive. Using the fact that $\Delta$ is a degree-two polynomial of $I_3$, one obtains the conditions ${I_1}^2 - 3 I_2 > 0$ and
$$
{-1} < \frac{27 I_3 - 9 I_1 I_2 + 2 {I_1}^3}{2 ({I_1}^2 - 3 I_2)^{3/2}} <1 \, ,
$$
to ensure that all eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$ are real. Their expression in terms of the invariants $I_1$, $I_2$, $I_3$ can be deduced from the polar decomposition of the Cardano formulas:
$$
\lambda_k = \frac{I_1}{3} + 2\sqrt{\frac{-p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{-q}{2}\sqrt{\frac{27}{-p^3}}\right) + \frac{2k\pi}{3}\right) ,
$$
for $k\in\lbrace 1,2,3\rbrace$.
It remains to compute the derivatives $\partial\lambda_k/\partial I_\ell$.
Alternatively, one might have computed the inverse of the Jacobian matrix $$
\frac{\partial \boldsymbol{I}}{\partial \boldsymbol{\lambda}} = \left(\frac{\partial I_i}{\partial \lambda_j}\right) = \begin{pmatrix}
1 & 1 & 1\\
\lambda_2+\lambda_3 & \lambda_1+\lambda_3 & \lambda_1+\lambda_2\\
\lambda_2\lambda_3 & \lambda_1\lambda_3 & \lambda_1\lambda_2
\end{pmatrix}
$$
whose entries are the requested partial derivatives expressed in terms of the $\lambda_k$. Their expression above allows to conclude.
