If we have local morphisms $A\to C$ and $B \to C$ of local artinian rings then the product $A \times_C B$ need not be local artinian anymore. To see this, take $A = \mathbb{C}(x)[\epsilon]$, $B = \mathbb{C}(y)[\delta]$ and $C = \mathbb{C}(x,y)$, with the obvious ring maps sending $\epsilon$ and $\delta$ to $0$. (See comment here.)
I'm interested what happens if we moreover assume the residue fields of $A$, $B$, $C$ to be isomorphic to a fixed field $k$, and in addition the maps $A \to C$ and $B \to C$ induce isomorphisms on residue fields. the result is true if we also assume that our rings are algebras but I'd rather not..!
Here is an attempt at a counterexample. It is based on the example given here: Tag 06S4
Lemma: Let $k$ be a ring, let $A$ be a $k$-algebra, let $D : A \to A$ be a $k$-derivation of $A$. Then the map $f_{D} : A \to A[\epsilon]$ sending $a \mapsto a + D(a) \cdot \epsilon$ is a $k$-algebra homomorphism.
Example: Let $k$ be a field of characteristic $0$, set $L := k(x)$. Set $A := L[\epsilon]$ and $B := L$ and $C := L[\epsilon]$. Let $A \to C$ be the map $a_{1} + a_{2} \cdot \epsilon \mapsto f_{\partial / \partial x}(a_{1}) = a_{1} + (\partial / \partial x)(a_{1}) \cdot \epsilon$ and let $B \to C$ be the map $b \mapsto b + 0 \cdot \epsilon$. Then an element $(a_{1} + a_{2} \cdot \epsilon , b)$ of $A \times B$ is contained in $A \times_{C} B$ if $a_{1} + (\partial / \partial x)(a_{1}) \cdot \epsilon = b$, which means $a_{1} = b$ and $(\partial / \partial x)(a_{1}) = 0$; since $k$ has characteristic $0$, the latter condition is equivalent to $a_{1} \in k$. Hence $A \times_{C} B \simeq k + L[\epsilon]$, which is not Artinian because $L$ is not a finite-dimensional $k$-module. (Note that $A,B,C$ are $L$-algebras but the above map $A \to C$ is not a map of $L$-algebras (it is only a map of $k$-algebras).)
Remark: (From here.) If we only impose the condition that the residue field extensions induced by $A \to C$ and $B \to C$ are finite extensions, then we have the simpler example $C := k(x)$ and $A := k(x^{2})[\epsilon]$ and $B := k((x+1)^{2})$ in which case $A \times_{C} B = k + k(x^{2}) \cdot \epsilon$.