Answer
a) $\nabla f(x,y)=\frac{1}{y} \hat{i}-\frac{x}{y^2}\hat{j}$
b) $<1,-2>$
c) $-1$
Work Step by Step
a) The gradient of $f(x,y)$ is the vector created by the partial derivative of $f$ with respect to $x$ and the partial derivative of $f$ with respect to $y$.
$ f_{x} = 1/y $
($1/y$ can be pulled out as a constant and the derivative of $x$ is $1$.)
$ f_{y} = -x/y^2$
($x$ can be pulled out as a constant and $1/y$ can be represented as $y^{-1}$ and then we solve using the power rule.)
Set these two partial derivatives up as a vector to get the gradient $f(x,y)$ as:
$\nabla f(x,y)=\frac{1}{y} \hat{i}-\frac{x}{y^2}\hat{j}$
b) We plug in the point $(2,1)$ for $x$ and $y$ in the gradient of $f(x,y)$ to get:
$<1,-2>$
c) To find the rate of change of $f(x,y)$ at the given point in the direction of the vector u, you set up the formula:
$Duf=f_{x}*a +f_{y}*b $
where a and b are the x and y components of the vector u given as $=<3/5, 4/5>$ (u must be a unit vector and in this case it already is). Plugging in the information given and what we solved in part b we get:
$ Duf=1*(3/5)+ (-2)(4/5)$ = $-5/5=-1$
