I'm not quite sure what you are saying with your conditions.
Since \frac{\partial}{\partial x}(u(y))=0=\frac{\partial}{\partial y}(v(x))∂∂x(u(y))=0=∂∂y(v(x)) for all xx and yy, the first of the Cauchy-Riemann equations doesn't put any conditions on the functions in the form given.
Since \frac{\partial}{\partial y}(u(y))=u'(y) and -\frac{\partial}{\partial x}(v(x))=-v'(x), the second of the Cauchy-Riemann equations implies that u'(y)=-v'(x) for all x and y. Since the LHS is a function of y and the RHS is a function of x, it follows that u'(y)=-v'(x)=A for some (real) constant A.
Integration now implies u(y)=Ay+B and v(x)=-Ax+C for some (real) constants B and C.
Therefore, functions of the form f(z)=(Ay+B)+i(-Ax+C) are the functions of the form f(z)=u(y)+iv(x) that are analytic. This corresponds to functions w=f(z) of the form f(z)=-iAz+(B+iC) where A, B, and C are real.
The geometric effect of such a function is to dilate a given complex number/vector z by |A|, reflect it through the origin if A<0, rotate it clockwise by 90^{\circ} (because of multiplying by -i), translate horizontally by B units (right if B>0 and left if B<0), and then translate vertically by C units (up if C>0 and down if C<0).