You need some conditions on K to make sense of the question. For instance, y ⟼ y2 is not surjective when K = ℝ and n = 2, because you cannot reach [[-1,0],[0,-2]]. Of course, you might have meant surjectivity in the sense of algebraic groups rather than in the sense of set-theoretic groups. But I think that that just amounts to asking for K to be algebraically closed. There is another problem when K has characteristic p. In this case y ⟼ yp is not surjective; its image is only the diagonalizable matrices.
The conjecture as stated seems plausible when K = ℂ. Or if K has positive characteristic and is algebraically closed, you could perhaps ask for a Zariski-dense image.
Updates:
Re Tom's comment: I think that Harald's question is clear enough, even though I agree that one has to work a little to see what he means. He must mean that w is an element of the free group in the letters x and y, of course. The second objection is also not essential. The question can be phrased as: For each w, what are the Zariski topology properties of the set of all x for which y ⟼ w(x,y) is surjective? I believe that Chevalley's theorem on constructible images tells you that the set of such x is constructible, so it either contains a Zariski open or it is disjoint from a Zariski open.
Re Harald's comment concerning if-and-only-if conditions. You have probably thought of this, but here goes anyway. If your derivative is non-singular for any z, then the w-map is Zariski dense. Of course, there are interesting maps in algebraic geometry that are Zariski dense but not surjective. You seem to suggest that that cannot happen here, but I do not know how to prove it.
