One solution would be to apply Russell's analysis of definite descriptions. Statements with n-place functions, f, can be rewritten as more complicated statements involving (n+1)-place functional relations, F (see NOTE 1).
For 1-place functions this looks like:
f(x) = y iff Fxy ∧ ∀z(Fxz → z = y))
For 2-place functions (like division) this looks like:
f(x,y) = z iff Fxyz ∧ ∀w(Fxyw → w = z))
For division in particular, this looks like:
x/y = z iff div(xyz) ∧ ∀w(div(xyw) → w = z))
Since the first conjunct, div(xyz) is false for every choice of z, no matter the choice of x and y, the entire statement will always be false. Intuitively, if a statement that uses a functional relation (like "the present king of France") implies the existence of something that doesn't actually exist, then (since any statement that implies a falsehood is false) such statements will be false.
"The present king of France is bald" is false, because there is no present king of France, and there would need to be in order for that statement to be true. "1/0=666" can be read as "the value of 1/0 is 666" which is also false, because there is no value of 1/0 and there would need to be for that statement to be true.
Of course, this is only one solution (one that uses "classical" logic). You can also use a three valued logic or one that allows statements with no truth values.
NOTE 1: The way this works in general is n-place functions f(x₁,x₂,...,xₙ) are replaced with (n+1)-place relations F(x₁,x₂,...,xₙ,y) and an axiom that guarantees existence and uniqueness for each combination of x₁,x₂,...,xₙ if f is a function (or only a uniqueness axiom if f is a partial function on the domain like division, or in discourse like philosophy where the domain is unclear):
∀x₁∀x₂...∀xₙ∀y∀z((F(x₁,x₂,...,xₙ,z) ∧ F(x₁,x₂,...,xₙ,z)) → y = z)
Then, every statement that refers to the value of f gets rewritten to state the existence of a y determined by F, which is guaranteed to be unique, and to refer to that y instead:
P(... f(x₁,x₂,...,xₙ) ...) becomes ∃y(F(x₁,x₂,...,xₙ,y) ∧ P(... y ...))