Let's start with what we know, which is that the probability distribution function (pdf) for this problem is:
$f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ (and 0 otherwise)
To obtain the joint density function (since the observations are independent), we simply take the product of the individual pdfs:
$f(x_1,x_2,...,x_n)=\prod_{i=1}^n f(x_i)=\prod_{i=1}^n \lambda e^{-\lambda x_i}$
(in your example, we have 3 "x's" and so the joint pdf is:)
$f(.1,.5,.9)=\lambda^3e^{-1.5\lambda}$
Then, take the log of this function:
$=n\ln\lambda-1.5\lambda$
Now differentiate with respect to lambda:
$\frac{\partial}{\partial\lambda}\ln L(\lambda)=\frac{n}{\lambda}-1.5$
and solve for lambda (which we will now denote $\hat{\lambda}$ because it is an estimator (specifically the MLE).
Here you get $\hat{\lambda}=2$