What happens to $w=\frac{x^2y^3}{z^4}$ , if $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively? If $x,y,z$ increase by $\%1$ , $\%2$ , $\%3$ respectively , then $w=\dfrac{x^2y^3}{z^4}$ ............. approximately.
$1)\ \%\ 3\text{ decrease}$
$2)\  \%\ 4\text{ decrease}$
$3)\ \%\ 3\text{ increase}$
$4)\ \%\ 4\text{ increase}$
I denote the new $w$ by $w'$,
$$w'-w=\left( \frac{1.01^2.1.02^3}{1.03^4}-1\right)\times w=\left(\frac{101^2\times102^3}{103^4\times100}-1\right)\times w$$
From here should I evaluate $\dfrac{101^2\times102^3}{103^4\times100}-1$ by hand (calculator is not allowed) or there is a quicker method to get to the correct answer?

You can use partial elasticities, also know as condition numbers:
\begin{align*}
\varepsilon_w \approx & \frac{x w'_x}{w} \varepsilon_x+\frac{y w'_y}{w} \varepsilon_y + \frac{z w'_z}{w} \varepsilon_z\\
= & 2\varepsilon_x + 3 \varepsilon_y - 4 \varepsilon_z= 2 + 3\times 2-4 \times 3 = -4
\end{align*}
So, the answer is an approximate decreasing by $\text{4%}$.
This comes from Taylor's formula:
\begin{align*}
f(\tilde x) \approx f(x) + f'(x)(\tilde x -x) \Rightarrow  \\
f(\tilde x) - f(x) \approx f'(x)(\tilde x -x) \Rightarrow \\
\frac{f(\tilde x) - f(x)}{f(x)} \approx  \frac{f'(x)(\tilde x -x)}{f(x)} \Rightarrow\\
\frac{f(\tilde x) - f(x)}{f(x)} \approx  \frac{xf'(x)}{f(x)} \frac{\tilde x -x}{x}
\end{align*}
