Reducing lambda expression to normal form
$\lambda$-calculus is a rewriting system, but it's power to express all partial recursive functions makes it equivalent to Turing machines, in the sense that it is a universal model of computation. Means of computation in $\lambda$-calculus are conversions and reductions, of which the main ones are $\beta$-reduction and $\alpha$-conversion. You can learn a lot about $\lambda$-calculus on Wikipedia alone.

$\alpha$-conversion
Take for example $\lambda$-expressions $(\lambda x. x)$ and $(\lambda y. y)$. We say they are $\alpha$-equivalent because they clearly represent the same function. In short, we can rename variables (and sometimes have to).

$\beta$-reduction
This relation is the one that "gives life" to $\lambda$-expressions and reduces them to normal forms. We use $\beta$-reduction on terms of the form $(\lambda x. M)\ N$, i.e. when applying abstraction to some term $N$, and we call that construction a redex. What this expression becomes after $\beta$-reduction is written $M[x:=N]$ and it means that we replaced all free occurrences of $x$ in $M$ with $N$. For example:
$$ (\lambda x. x\ x)\ N \rightarrow_\beta N\ N$$

We will be reducing leftmost outermost redex in each step. This strategy garantees that we will get to normal form if it exists.  

Now let us dive into this...
Let's find a leftmost outermost redex (underlined) in your expression:
  $$\underline{\Bigl(\lambda n.\lambda m.\lambda f.\lambda x. n\ (m\ f)\ x\Bigr)\ \Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)}\ (\lambda f. \lambda x. f\ x)$$

We have $\Bigl(\lambda n.\lambda m.\lambda f.\lambda x. n\ (m\ f)\ x\Bigr)$, which is an abstraction (it is of the form $\lambda n. M$), that is applied to $\Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)$, so ($\beta$-reduction) we have to replace all free occurences of $n$ in $M$ with $\Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)$. We get $\biggl(\lambda m.\lambda f.\lambda x. \Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)\ (m\ f)\ x\biggr)$, so entire expression is now (where I already underlined next outermost redex):
  $$\underline{\biggl(\lambda m.\lambda f.\lambda x. \Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)\ (m\ f)\ x\biggr)\ (\lambda f. \lambda x. f\ x)}$$
We now have to substitute $(\lambda f. \lambda x. f\ x)$ for $m$ in $\biggl(\lambda f.\lambda x. \Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)\ (m\ f)\ x\biggr)$ and get
  $$\lambda f.\lambda x. \underline{\Bigl(\lambda f. \lambda x. f\ (f\ x)\Bigr)\ \Bigl((\lambda f. \lambda x. f\ x)\ f\Bigr)}\ x$$
Now just keep reducing
  $$\lambda f.\lambda x. \underline{\Bigl(\lambda x. ((\lambda f. \lambda x. f\ x)\ f)\ (((\lambda f. \lambda x. f\ x)\ f)\ x)\Bigr)\ x}$$
  $$\lambda f.\lambda x. \Bigl(\underline{(\lambda f. \lambda x. f\ x)\ f}\Bigr)\ \biggl(\Bigl((\lambda f. \lambda x. f\ x)\ f\Bigr)\ x\biggr)$$
  $$\lambda f.\lambda x. \underline{(\lambda x. f\ x)\ \biggl(\Bigl((\lambda f. \lambda x. f\ x)\ f\Bigr)\ x\biggr)}$$
  $$\lambda f.\lambda x. f\ \biggl(\Bigl(\underline{(\lambda f. \lambda x. f\ x)\ f}\Bigr)\ x\biggr)$$
  $$\lambda f.\lambda x. f\ \Bigr(\underline{(\lambda x. f\ x)\ x}\Bigr)$$
  $$\lambda f.\lambda x. f\ (f\ x)$$
And now, if you are still alive, remember $\alpha$-equivalence. We can change the names of variables, say $f$ to $y$ and $x$ to $z$, and obtain your wanted solution.

What have we computed?
We computed that $2 \cdot 1 = 2$, in $\lambda$-calculus using Church numerals:


$c_0 = \lambda f. \lambda x. x$
$c_1 = \lambda f. \lambda x. f\ x$
$c_2 = \lambda f. \lambda x. f\ (f\ x)$
$\vdots$  


Expression $\lambda n.\lambda m.\lambda f.\lambda x. n\ (m\ f)\ x$ is the multiplication operator, and other two expressions (on which this operator was applied) are Church numerals for $2$ and $1$.