Derive the generating function: $\frac1{(1-x)^3}= \sum_{n=0}^\infty \binom{n+2}{2}x^n$

Note that, indeed, $\dfrac 1 {1-x} = \sum \limits _{n=0} ^\infty x^n$. As you tried to do, let us differentiate this twice. The first time you get

$$\frac 1 {(1-x)^2} = \sum \limits _{n=1} ^\infty n x^{n-1}$$

(note that the sum starts from $1$ now, not from $0$). Differentiating once more, you get

$$\frac 2 {(1-x)^3} = \sum \limits _{n=2} ^\infty n(n-1) x^{n-2}$$

and, if you change the summation variable according to $m = n-2$, you get

$$\frac 2 {(1-x)^3} = \sum \limits _{m=0} ^\infty (m+2)(m+1) x^m\;,$$

which is your desired result because

$$\binom {m+2} 2 = \frac {(m+2)!} {m! \ 2!} = \frac {(m+2)(m+1)} 2\;.$$

Take the derivative of both sides

$$\frac{d}{dx} \left (\frac{1}{(1-x)^2} \right) = \frac{d}{dx} \left ( \sum_{n=0}^{\infty} n x^{n-1} \right)$$

$$-(-2)\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} n (n-1) x^{n-2}$$

$$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} \frac{n (n-1)}{2} x^{n-2} = \sum_{n=0}^{\infty} \frac{(n +2) (n+1)}{2} x^{n}.$$

Look at the definition of the binomial coefficient and you'll see that this is the same as the desired formula.