This is a homework question, so I'd appreciate hints (or perhaps explanations of concepts I've not properly digested)
Anyhow: This is exercise 1.3.6 in Weibel's book on homological algebra. Let $0 \to A \to B \to C \to 0$ be an exact sequence of double complexes of modules. Show that there is a short exact sequence of total complexes, and conclude that if Tot(C) is acyclic, then $Tot(A) \to Tot(B)$ is a quasi-isomorphism.
The last part of the exercise is clear. If Tot(C) is acyclic, then the long exact sequence is of the form
$\ldots \to H_{i+1}C(=0) \to H_i(A) \to H_i(B) \to H_i(C)(=0) $
so the induced morphism on homology is an isomorphism.
The first part of the question is unclear, however. The definition of an exact sequence of double complexes is not explicitly stated, but I assume it is such that $0 \to A_{ij} \to B_{ij} \to C_{ij} \to 0$ is exact for all i,j and everything commutes.
Let $\alpha:A \to B$ be a morphism of double complexes. The induced morphism between the total complexes, $\alpha^*: Tot(A) \to Tot(B)$, is then defined, I'd assume, as $\alpha^*=d_B^h \alpha + d_B^v \alpha$ (where $d_B^h$ and $d_B^v$ denotes the horizontal and vertical differentials, respectively). The problem is how I go about showing the induced sequence is exact.
"EDIT:" After some thought, I guess a good first step would be to show that $\beta^* \circ \alpha^* = 0$, which shouldn't be too difficult. (where $\alpha^*,\beta^*$ denotes the induced morphisms of total complexes)
Edit2: I've clarified my notation a bit.