If I have a function $$f(x,y)=\sqrt{x^2+y^2}$$ with error in $x$ be $\Delta x$ and error in $y$ be $\Delta y$, then how do we calculate ${\Delta f}$?
I know if we have $$f(x)=x^n$$, then I at least that $\frac{\Delta f}{f}=|n|\frac{\Delta x}{x}$. I have no idea how to proceed.
On
The error is typically approximated using the differential; i.e. the first-order term of the Taylor series:
$$ \Delta f \approx df(\Delta x, \Delta y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y.$$
For your particular $f$ we have $\partial f / \partial x = x/f, \partial f / \partial y = y/f$.
The partial derivatives with respect to $x$ and $y$ play a similar role as the derivative does in the one-variable case. We have $$\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} \quad\text{and}\quad \frac{\partial f}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}.$$ Thus the error $\Delta f$ is given approximately by $$\frac{x}{\sqrt{x^2+y^2}} \Delta x +\frac{y}{\sqrt{x^2+y^2}}\Delta y.$$
A more intuitive version of the same thing is to write $$f^2=x^2+y^2.$$ The error in $f^2$ is approximately $2f\Delta f$, where $\Delta f$ is the error in $f$. Similarly, the error in $x^2$ is approximately $2x\Delta x$, and the error in $y$ is approximately $2y\Delta y$. Thus $$2f\Delta f\approx 2x\Delta x+2y\Delta y.$$