Is this a counterexample to a conjecture about independent domination in cartesian graph products? VIZING’S CONJECTURE: A SURVEY AND RECENT RESULTS (2009) by Bostjan Bresar , Paul Dorbec , Wayne Goddard , Bert L. Hartnell , Michael A. Henning , Sandi Klavzar , Douglas F. Rall 
   p.25:

Conjecture 9.6. For all graphs $G$ and $H$,
  $$\gamma(G \square H) \ge \min\{i(G)\gamma(H), i(H)\gamma(G)\}$$

where $\square$ is the cartesian product of graphs and $i(G)$ is the independent domination number.
For cartesian squares it is $$\gamma(G \square G) \ge \gamma(G) i(G)$$
According to sage and my verification square of the graph on 7 vertices $[0 \ldots 6]$ with edges
$$[(0, 4), (0, 5), (0, 6), (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6)]$$
appears a counterexample to Conjecture 9.6 (note that vertex 3 is disconnected).
Computation:
sage: G=Graph(':Fo@I@I@J') #from sparse6
sage: Gs=G.cartesian_product(G)
sage: (Gs.dominating_set(value_only=True),G.dominating_set(value_only=True),G.dominating_set(value_only=True,independent=True) )
(11, 3, 4)

Verification:
Since the order is only $7$, $\gamma(G)=3$ and $i(G)=4$ were verified by enumerating
all subsets of the vertices. For $\gamma(G \square G)=11$ the dominating set returned by sage
was verified and it is an upper bound for the correct value.
I well might have misunderstood the conjecture.

Is the above graph a counterexample of Conjecture 9.6?

Adding disconnected vertices to $G$ gives more counterexamples.
 A: To summarize a bit:


*

*$\gamma(G)$ is defined as the usual domination number of a graph $G$

*$i(G)$ is defined as the smallest cardinality of a dominating set that is also an independent set


Your graph $C$ is the disjoint union of $K_{3,3}$ and $K_1.$
Clearly $\gamma(C) = 3$ and $i(C) = 4.$ For disjoint graphs $G,H$ we have $$(G \cup H) \square (G \cup H) = (G \square G) \cup (G \square H) \cup (H \square G) \cup (H \square H)$$
which gives for $C = K_{3,3} \cup K_1$
$$ C \square C = K_{3,3} \square K_{3,3} \cup K_{3,3} \cup K_{3,3} \cup K_1.$$
And thus $$\gamma(C \square C) = \gamma(K_{3,3} \square K_{3,3}) + 2\gamma(K_{3,3})+\gamma(K_1) = 11.$$
This would indeed imply that $\gamma(C \square C) = 11 < \gamma(C)i(C) = 12.$ Making the conjecture false for disconnected graphs. 
Edit. In this paper the authors construct an infinite family of graphs that are a counterexample to the claim of conjecture 9.6. The constructed family of graphs is disconnected but they remark it can be made connected. 
A: Since commenters asked about a connected counterexample, here is the connected counter example
from the paper
 Jernej mentions.
p. 2. A $k$-leaf star is a central vertex with $k$ vertices connected to it. A $2k$-leaf-two star
is a graph from two $k$-leaf stars with their central vertices connected by an edge.
Define $G_k$ as the disjoint union of $2k^2$-leaf-two star and $k$ $K_2$s.
$\gamma(G_k)=k+2$ and $i(G_k)=k^2+k+1$.
The disconnected counterexample is $F = G_k \square G_k$ with given dominating set
$\gamma(F) \le 12k^2+8k+4$.
To make connected counterexample $G_k'$, add root vertex $r$ to $G_k$, connect it to
one center of the two star and one vertex of each $K_2$.
$\gamma(G_k' \square G_k') \le 16k^2+12k+9$.
Because of the constants the smallest disconnected counterexample guaranteed by
the construction is on $266$ vertices.
