I don't know how to star solving this problem, I'm new in calculus..
f(x+y) - f(x) = sec^2 x tan y / 1 +tan x tan y
As Ackbeet said, this is a matter of a trig identity and you can reduce to sine and cosine.
$\displaystyle tan(x+y)= \frac{sin(x+y)}{cos(x+y)}= \frac{sin(x)cos(y)+ cos(x)sin(y)}{cos(x)cos(y)- sin(x)sin(y)}$
Now, divide both numerator and denominator by cos(x)cos(y)
$\displaystyle tan(x+y)= \frac{\frac{sin(x)}{cos(x)}+ \frac{sin(y)}{cos(y)}}{1- \frac{sin(x)}{cos(x)}\frac{sin(y)}{cos(y)}}$
$\displaystyle = \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}$
So that $\displaystyle tan(x+y)- tan(x)= \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}- tan(x)$
$\displaystyle = \frac{tan(x)+ tan(y)}{1- tan(x)tan(y)}- \frac{tan(x)- tan^2(x)tan(y)}{1- tan(x)tan(y)}$
Note that, as skeeter said, you have the sign in the denominator wrong.