In the formula [called the Lens Equation] if I have to find maximum permissible error in $f$ (from the graph) which is the correct method?
$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$
Method 1:
$${df}=f^2\left(\frac{dv}{v^2}+\frac{du}{u^2}\right)$$
Method 2:
$${f}=\frac{uv}{u+v}$$
$$\ln(f)=\ln(uv)-ln(u+v)$$
$$\implies \frac{df}{f}=\frac{du}{u}+\frac{dv}{v}+\frac{d(u+v)}{u+v}$$
Both these methods give different results. Which is the correct one?
In doing error analysis, we're always looking for the worst-case scenario where the errors add up rather than cancel. To this end, one is best off using specific signs of the quantities while differentiating and use absolute values at the end to obtain the error estimates. With this in mind, the two methods above correspond to:
Method 1: $$\displaystyle \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\implies -\frac{df}{f^2}=\frac{dv}{v^2}-\frac{du}{u^2}\implies \left|\frac{\delta f}{f^2}\right| \geq \left|\frac{\delta v}{v^2}\right|+\left|\frac{\delta u}{u^2}\right|$$ where in the last line I've taken absolute values and replaced diffentials with finite errors $\delta u,\delta v$. Note the change of signs in the last step, consistent with the fact that the errors in $u,v$ only add up when they have opposite signs.
Method 2: \begin{align} f=\frac{uv}{u-v} \implies \log f&= \log u+\log v-\log(u-v)\\ \implies \frac{df}{f}&=\frac{du}{u}+\frac{dv}{v}-\frac{du-dv}{u-v}\\ &=-\frac{v}{u-v}\frac{du}{u}+\frac{u}{u-v}\frac{dv}{v}\\ &=-\frac{f}{u}\frac{du}{u}+\frac{f}{v}\frac{dv}{v}\end{align} which rearranges to the same equality as in Method 1. Hence the two approaches agree.