Fastest increase of a function For the function $f(x,y,z) = \frac{1 }{ x^2+y^2+z^2}$ what is the direction of the fastest increase at $(1,1,1)$? 

In the direction of the gradient of $f$ evaluated at $(1,1,1)$. 
This is a general fact concerning   functions $f=f(x,y,z)$:  The direction in which a differentiable function $f$ increases most rapidly at the point $(a,b,c)$ is in the direction of $\nabla f(a,b,c)$.
You'll need to find the general formula for $\nabla f(x,y,z)$ first.  Then simply evaluate $\nabla f(1,1,1)$. The solution is below, but it would be to your benefit to try the computations before looking at it.



Solution:
The first thing you have to do is compute the gradient of $f$ in general; that is compute $\nabla f(x,y,z)$:
$$\eqalign{
\nabla f(x,y,z) &=f_x(x,y,z)\,{\bf i} +f_y(x,y,z)\,{\bf j} + f_z(x,y,z)\,{\bf k} \cr
&={\partial \over\partial x} {1\over x^2+y^2+z^2}\,{\bf i} 
+{\partial \over\partial y} {1\over x^2+y^2+z^2} \,{\bf j}
+{\partial \over\partial z} {1\over x^2+y^2+z^2}\,{\bf k} \cr
&= -(x^2+y^2+z^2)^{-2}\cdot 2x\,{\bf i}
 -(x^2+y^2+z^2)^{-2}\cdot 2y\,{\bf j}
 -(x^2+y^2+z^2)^{-2}\cdot 2z\,{\bf k}\cr
&={ -2x\over (x^2+y^2+z^2)^{2}}\,{\bf i}
+ { -2y\over (x^2+y^2+z^2)^{2}}\,{\bf j}
+ { -2z\over (x^2+y^2+z^2)^{2}}\,{\bf k}.
}
$$
Now evaluate the gradient at the point $(1,1,1)$:
$$\eqalign{
\nabla f(1,1,1)
&={ -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf i}
+{ -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf j}
+ { -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf k}\cr
&={-2\over9}\,{\bf i}+{-2\over9}\,{\bf j}+{-2\over9}\,{\bf k} .
}
$$
