Partial derivatives

If sin (2x+4y+z) = 0 , find the first partial derivatives $$\frac{dz}{dx}$$ at the point (0,0,0)

A.) $$\frac{dz}{dx}(0,0,0) =$$ _________________

isnt this saying get the derivative of z, respect to x? i'm just kinda confuse since the variable 'z' is also in the problem.

well i got the derivative of that function with respect to x and got 2*cos(2x+4y+z), plugged in 0,0,0 and got 2*cos(0), which is wrong. the answer should be -2*cos(0), where did the negative sign come from?

$$\sin(2x+4y+z(x,y))=0$$