Show that the product of two convex functions $f,g$ is convex, when both $f,g$ are positive and nondecreasing

Yes, positive means the ranges of the functions are contained in $(0,\infty)$. It is good enough to assume that the ranges are in $[0, \infty)$.

Hint: Just apply the definition of convexity to $f$ and $g$. The question reduces to the following:

$$ \theta^{2}f(x)g(x)+(1-\theta)^{2}f(y)g(y)+\theta (1-\theta) [f(x)g(y)+g(x)f(y)]$$ $$ \leq \theta (f(x)g(x)+(1-\theta) f(y)g(y).$$

Some simple algebraic manipulation reduces this to

$$f(x)g(y)+g(x)f(y) \leq f(x)g(x)+f(y)g(y).$$

This can be written as $$f(x)[g(y)-g(x)] \leq f(y)[g(y)-g(x)].$$

Can you prove this by considering the cases $x \leq y$ and $x >y$?

[In the first case use the fact that $a \leq b$, $c \leq d$ with $a,b,c,d \geq 0$ implies $ac \leq bd$ because $bd-ac =b(d-c)+c(b-a)\geq 0$].