If you need the final expression to be a CNF, then I don't think you can simplify much. The minimal number of conjuncts that I can see is three, versus four in the original formula. Also, it is difficult to do the simplification without a Karnaugh map.

Below, the first two steps split some conjuncts into two, and the rest of the steps recombine them in new ways. I write comments in curly braces.

xyw' + xzw + y'zw + z'w' = {xyw' = xy(z + z')w' = xyzw' + xyz'w'}

xyzw' + xyz'w' + xzw + y'zw + z'w' = {xzw = x(y + y')zw = xyzw + xy'zw}

xyzw' + xyz'w' + xyzw + xy'zw + y'zw + z'w' = {xyz'w' + z'w' = (xy + 1)z'w' = z'w'}

xyzw' + xyzw + xy'zw + y'zw + z'w' = {xy'zw + y'zw = (x + 1)y'zw = y'zw}

xyzw' + xyzw + y'zw + z'w' = {xyzw' + xyzw = xyz(w + w') = xyz}

xyz + y'zw + z'w'

Edit: It is recommended to post questions about Boolean algebra in the Discrete Math subforum.