Calculate$\operatorname{Var}(Y |X = 2)$ for a given pdf. 
A machine has two components and fails when both components fail. The
  number of years from now until the first component fails, X, and the
  number of years from now until the machine fails, Y , are random
  variables with joint density function
  $$f(x,y)=\begin{cases}\frac1{18}e^{-(x+y)/6}&\text{if }0<x<y\\
0&\text{otherwise}\end{cases}$$
  Calculate $\operatorname{Var}(Y |X = 2)$.

My answer: I can notice that the ftp is a product of two exponentials rv, with $\lambda= \frac16$. And the variance of an exponential is $\frac{1}{\lambda^2}$. Then the variance of this pdf is $1/(1/6)^2=36$. Is that right? The book said it is $36$, but I'm not sure if my argument is correct.
My question: How to calculate $\operatorname{Var}(Y |X = 2)$?
 A: I don´t know if your argument is right. I have used the definition of the conditional variance.
$Var(Y|X)=\int_{-\infty}^{\infty} y^2\cdot h(y|x)\ dy- \left[ \int_{-\infty}^{\infty} y\cdot h(y|x)\ dy\right] ^2$
$h(y|x)=\frac{f(x,y)}{f_X(x)}$
$f_X(x)=\int_x^{\infty} \frac{1}{18}\cdot e^{-(x+y)/6} \ dy=\frac{1}{3}e^{-x/3}$
$h(y|x)=\frac{\frac{1}{18} e^{-(x+y)/6}}{\frac{1}{3}e^{-x/3}}=\frac{1}{6}e^{(x-y)/6}$
$h(y|2)=\frac{1}{6}e^{1/3-y/6}$
$\int_{2}^{\infty} y^2\cdot h(y|2) \ dy=\int_2^{\infty} y^2\cdot 1/6\cdot e^{1/3-y/6} \ dy=100$
$\int_{2}^{\infty} y\cdot h(y|2) \ dy=\int_2^{\infty} y \cdot 1/6\cdot e^{1/3-y/6} \ dy=8$
A: You've observed that $f_{X,Y}(x,y)=\tfrac 1{18}\mathsf e^{-x/6}\mathsf e^{-y/6}$ is a product of two exponential functions and suggest this means it is the joint of two independent exponential distributions.
Let us test this.
$$\begin{align}f_X(x) & = \int_x^\infty \tfrac 1{18} \,\mathsf e^{-x/6}\,\mathsf e^{-y/6}\operatorname dy \;\mathbf 1_{0\le x}\\[1ex] & = \tfrac 1 3\mathsf e^{-x/3}\;\mathbf 1_{0\le x}\\[2ex] f_Y(y) & = \int_0^y \tfrac 1{18} \,\mathsf e^{-x/6}\,\mathsf e^{-y/6}\operatorname dx \;\mathbf 1_{0\le y}\\[1ex] & = \tfrac 1 3\left(\mathsf e^{-y/6}-\mathsf e^{-y/3}\right)\;\mathbf 1_{0\le y}\\[2ex] \therefore f_{X,Y}(x,y) \;& {\large\neq}\;f_X(x)\cdot f_Y(y) & \color{red}{\mathcal X}\end{align}$$
So, nope.

However, your instincts were not completely off track.   Let $Z=Y-X$.
$$\begin{align}f_{X,Z}(x,z) & = f_{X,Y}(x,x+z)\\[1ex] & =\tfrac 1{18}\mathsf e^{-x/6}\mathsf e^{-(z+x)/6} \;\mathbf 1_{0\le x\le z+x}
\\[1ex] & =\tfrac 1{18}\mathsf e^{-x/3}\mathsf e^{-z/6} \;\mathbf 1_{0\le x}\,\mathbf 1_{0\le z}
\\[2ex]
f_X(x) & = \int_0^\infty \tfrac 1{18} \,\mathsf e^{-x/3}\,\mathsf e^{-z/6}\operatorname dz \;\mathbf 1_{0\le x}\\[1ex] & = \tfrac 1 3\mathsf e^{-x/3}\;\mathbf 1_{0\le x}\\[2ex] f_{Z}(z) & = \int_0^\infty \tfrac 1{18} \,\mathsf e^{-x/3}\,\mathsf e^{-z/6}\operatorname dx \;\mathbf 1_{0\le z}\\[1ex] & = \tfrac 1 6\mathsf e^{-z/6}\;\mathbf 1_{0\le z}\\[2ex] \therefore f_{X,Z}(x,z) \;& {\large = }\;f_X(x)\cdot f_{Z}(z)& \color{green}{\checkmark}\end{align}$$
So while $X,Y$ are not independent, the random variables $X, Z$ are, and further they have the exponential distributions you suspected.
Then $\mathsf {Var}(Y\mid X=2) \\[1ex] = \mathsf {Var}(2+Z\mid X=2) \\[1ex] = \mathsf {Var}(Z)$
Now, as $Z\sim\mathcal{Exp}(1/6)$ you can find your variance.
