f(z) can be expressed as u(x,y) , iv(x,y). Am I doing this right? So, $f(z) = z^2 + 4z - 6i$ and I need to express this as $u(x,y)$ , $iv(x,y)$. So, I plug in $z = x+iy$ and simplify.  
I am left with this:
$f(x+iy) = x^2 - y^2 + 2xy + 4x + 4iy -6i$.  
Now, how do I express this as $u(x,y)$ and $iv(x,y)$ 
The reason I am doing this is because I read that the derivative of $f(z)$, that is $f\prime(z)$ is $u\prime(x,y) + iv\prime(x,y)$.
 A: When you expand $(x+iy)^2$ you get $x^2-y^2+2ixy$, so you have a small typo in your question. Now put $u(x,y) = x^2-y^2+4x$ and $v(x,y) = 2xy+4y-6$. Then
$$f(x+iy) = u(x,y) + iv(x,y).$$
A: A comment on the statement: 
The reason I am doing this is because I read that the derivative of $f(z)$ , that is $f'(z)$  is $u'(x,y)+iv'(x,y)$.
The partial derivatives $f_x$ and $f_y$ of $f(z)$ in $z=x+iy$ are defined as 
$$f_x:=u_x+iv_x,$$
and
$$f_y:=u_y+v_y.$$
In your question you write $u(x,y)=x^2-y^2+4x$ and $v(x,y)=2xy+4y-6$.
The partial derivatives $f_x$ and $f_y$
exist, as both $u(x,y)$ and $v(x,y)$ are polynomials.
$f$ differentiable (I interpret this as your derivative of $f$ at $z$) 
at $z$ is a much stronger condition; in fact $f$ is differentiable at $z$ if the limit
$$\lim_{h\rightarrow o}\frac{f(z+h)-f(z)}{h}$$
exists (independently on the complex vector $h$ we choose!). Differentiability at $z$ implies the Cauchy Riemann equations $u_x=v_y$ and $u_y=-v_x$ for $u(x,y)$ and $v(x,y)$, where $z=x+iy$. Given $f$ as in your question, you can quickly check that it is differentiable at $z$.
