Sketch the function $y = {1 \over {{x^2}}}\ln x$ I don't know where to begin with this, the ${1 \over {{x^2}}}$ part of the function throws me off, how to do I go about this? How does one generally approach a question like this?
 A: It is indeed an interesting function. You can try finding a lot of points: select a range of $x$ values, and evaluate $$f(x) = \dfrac{\ln x}{x^2}$$ for each x.  Given enough data points, plotting them as you go along, you can try to "connect the dots" to see what you get.
You can also check for asymptotes, critical values, if any (differentiate!!) to find any extrema, if they exist, etc: that information will help reveal if and where the function peaks and/or bottoms out, where it is increasing, decreasing, etc. 
For example, if we take the limit of $f(x)$ as $x\to +\infty$,  $x \to 0^+$, e.g., and we will see some interesting things. 
Computing the derivative gives us $$f'(x) = \frac{1-2\ln x}{x^3}$$ So the derivative $f'(x) = 0$ at $x=e^{1/2}$, is positive for $x\lt e^{1/2}$ (hence increasing) and is negative, hence decreasing, when $x\gt e^{1/2}$. 
Maximum: $x=e^{1/2}$.
You can also "take a peak" to see what you're trying to approximate: [Disclaimer Wolfram Alpha's graph is taking, I assume, $\ln x$ in the numerator to be $\ln|x|$, in which case you'd have a symmetrical graph (the left portion mirroring the right-hand portion).

A: The function is only defined for positive $x$. For $0<x\lt 1$, $\ln x$ is negative, large negative near $x=0$. Dividing by $x^2$ makes things much larger negative.
For $x$ large positive, the $x^2$ at the bottom crushes puny $\ln x$. So for $x$ large, our function is positive but close to $0$.
We might note that $\frac{\ln x}{x^2}$ is $0$ at $x=1$, and positive for $x\gt 1$.
For more detail, we compute the derivative.  This is $\frac{1-2\ln x}{x^3}$. Thus the derivative is $0$ at $x=e^{1/2}$, positive for $x\lt e^{1/2}$, and negative for $x\gt e^{1/2}$. 
So our function reaches a maximum at $x=e^{1/2}$. The maximum value is $\frac{1}{2e}$.
Now we have enough information to draw a pretty good sketch. For fine detail about concavity, one can look at the second derivative. If you do that calculation, you will find that the curve is facing down up to $x=e^{5/6}$, and then facing up. This does not add much in the way of useful information in this case. However, an exam question might ask about concavity, since it is not too hard to deal with in this example. 
A: If you aren't supposed to be using a calculator/comptuer: plug in lots of values of $x$ for which you'll be able to easily approximate the value of $f(x)$, plot the resulting points $(x,f(x))$, and connect them with a reasonable-looking curve.
For example, for the function $f(x)=\frac{1}{x^2}\ln(x)$, we'll want to plug in values of $x$ for which we know the approximate value of $\ln(x)$. Some such $x$'s are powers of $e$. For example, let's say that $e\approx 2.7$. Then we can approximate
$$f(2.7)=\frac{1}{(2.7)^2}\ln(2.7)\approx \frac{1}{(2.7)^2}\cdot 1=\frac{1}{7.29}$$
If you are allowed to use a calculator/comptuer, just tell it to do it :)
Entering this code in Mathematica 
Plot[(1/x^2) Log[x], {x, 0, 5}, PlotRange -> {-1, 1}]

produces

A: $x^{-2}>1$ when $|x|<1$ and if you know the graph of $lnx$ then the guessing part should go easy. $lnx$ goes off to $-\infty$ as $x$ approaches $0$. Because of $x^{-2}$ in that region, it will shoot off to $-\infty$ at a quicker rate. Whereas for $x>1$ region, it should be noted that $x^{-2}$ has a growth rate larger than $lnx$ and hence $x^{-2}$ would manage to calm $lnx$ down. This knowledge coupled with the knowledge of maxima should be enough to give a good estimate for all $x>0$.
