Finding minimisation as the sum of 3 terms from K-Map I was given the following Boolean function:
F(x, y, z) = xyz + xy'z + xy'z' + x'yz + x'yz' + x'y'z'
And from it, I was able to produce the following truth table and K-Map (which I assume both are correct):
 
However, now I am asked to find a minimisation as the sum of the three terms in the function F(x, y, z) given above.  I'm really stuck here and unfortunately the lectures were of no help.
 A: The problem you are having is that your assumption that the Karnough map you created is correct was false. You have placed a $0$ in the $x'y'z'$ box where there should in fact be a $1$.

Once you have corrected this, follow the procedure bellow to minimize the function. 
The way to minimize the function, is to circle neighboring terms in the Karnaugh map which are true. By circling two terms which are next to another, in a function which is of three variables, you will produce a two variable term. 

As an example, the leftmost blue circle I made contains the term $y'z'$, as it is true regardless of the truth of $x$.
Once you have circled every box that is true, simply OR the terms together to produce the final expression. The expression the question requires should only contain three terms, so try to create three circles. I had success with $x'y+xz+y'z'$, but other combinations exist.

Could you explain the logic behind F(x,y,z) = y'z' + xz + x'y a little more? How do they tie to the circling you just did?  How did you decide that they were xz and x'y?

EDIT: Onur Ozbek asked a few question in the comments addressing how the circle placement is decided. I hope the following image clears things up.

If $F(x,y,z)$ is true outside the $z$ column, then it is true for $z'$. The same goes for the variables $x$ and $y$.
A: Alternative solution:
\begin{array}{ccccc}&x'y'&x'y&xy&xy'\\z&0&\color{red}{1}&\color{red}{1}&\color{blue}{1}\\z'&\color{orange}{1}&\color{orange}{1}&0&\color{blue}{1}\end{array}
$$\color{orange}{x'z'}+\color{red}{yz}+\color{blue}{xy'}$$
