Differentiability of eigenvalues of positive-definite symmetric matrices
In the open subset of $M_n(\mathbb{R})$ where the $\lambda_i$ are distinct, they are $C^{\infty}$ functions: this follows from the implicit function theorem.
On the other hand, when some eigenvalue has multiplicity $>1$ you don't get more than continuity. For example if $A=\begin{pmatrix} 0 & 1\\ 1 & t \end{pmatrix}$ the largest $\lambda_i$ is $\dfrac{1}{2}\left(t^2+2 +|t|\sqrt{t^2+4}\right)$, which is not differentiable (as a function of $t$) at $t=0$.
The keyword is the Cartan decomposition in the theory of symmetric spaces.
In short, when an eigenvalue is simple (its multiplicity is $1$) it is locally an analytic function. But when the eigenspace is degenerate (the multiplicity is greater than $1$), the eigenvalue function is not differentiable. The problem is essentially one of choosing branches: if you try to deform the identity matrix, there is no consistent way to say which of the resulting distinct eigenvalues after deformation is the eigenvalue that you should have kept track of.
Let $K = \mathrm{O}(n)$, and let $A$ be the group of diagonal matrices with positive entries. You then have $G=KAK$ and if $g=k_1 a k_2$ then the eigenvalues of $g^\dagger g$ are exactly the squares of the eigenvalues of $a$. The problem is that the decomposition is not unique: you can conjugate $a$ by a permutation matrix, and there will be problems when $a$ is fixed by a permutation matrix.
As mentionned by other answers, simple eigenvalues are $C^\infty$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of linear operator.
The first one is that each $\lambda_j$ is a Lipschitz function. This statement is still valid if you replace ${\bf Sym}_n({\mathbb R})$ by a subspace $E\subset{\bf M}_n({\mathbb R})$ with the property that the eigenvalues are always real.
The second one is that if $t\mapsto A(t)$ is a smooth curve in ${\bf Sym}_n({\mathbb R})$, then there is a labelling of the eigenvalues $t\in{\cal V}\rightarrow(\mu_1(t),\ldots,\mu_n(t))$ such that each $\mu_j$ is smooth. Mind that this labelling does not respect the order between eigenvalues when the multiplicities vary. Mind also that this becomes false if we replace a curve by a surface.