If $f$ and $g$ are real-valued differentiable functions in an open set $E \subset \mathbb{R}^n$, show that $$\nabla\left(\frac{1}{f}\right)=-\frac{\nabla f}{f^2}$$
So if I have $$ \nabla(\frac1f) = \left( \frac{\partial}{̛\partial x}\frac{1}{f}, \frac{\partial}{̛\partial y}\frac{1}{f}, \frac{\partial}{̛\partial z}\frac{1}{f} \right) $$ wouldn't this equal $$ \left(-\frac{\partial}{̛\partial x}\frac{1}{f^2}, -\frac{\partial}{̛\partial y}\frac{1}{f^2}, -\frac{\partial}{̛\partial z}\frac{1}{f^2}\right) = -\frac{\nabla f}{f^2}? $$ I'm taking the factor $-\frac{1}{f^2}$ just out which leaves me with $\nabla f.$
I'm a bit confused about the term $\left(-\frac{\partial}{̛\partial x}\frac{1}{f^2}, -\frac{\partial}{̛\partial y}\frac{1}{f^2}, -\frac{\partial}{̛\partial z}\frac{1}{f^2}\right)$ I'm thinking this the way that after differentiating with respect to $x$ I get $-\frac{\partial}{̛\partial x}\frac{1}{f^2}$ from $\frac{\partial}{̛\partial x}\frac{1}{f}$ and $f^2$ would still contain the other variables in this case up to $z$, but it could be probably any number one desires.
As noticed in the comments, by chain rule we have for any components
$$\frac{\partial}{\partial x_i} \left(\frac1f\right)=-\frac1{f^2}\frac{\partial f}{\partial x_i}$$
and therefore $\nabla\left(\frac{1}{f}\right)=-\frac{1}{f^2}\nabla f$.
Let consider as a concrete example $f(x,y)={x^2+y}$ then
$$\frac{\partial}{\partial x} \left(\frac1{x^2+y}\right)=-\frac{2x}{(x^2+y)^2}=-\frac1{f^2}\frac{\partial f}{\partial x}$$
$$\frac{\partial}{\partial y} \left(\frac1{x^2+y}\right)=-\frac{1}{(x^2+y)^2}=-\frac1{f^2}\frac{\partial f}{\partial y}$$