Let $x=1$ and $x+\delta x=1.04 \ \therefore \delta x = 0.04$

Also Let $y=3$ and $y+\delta y=3.01 \ \therefore \delta y=0.01$

Let $z=x^y \ \cdots \ \cdots (1)$

Differentiating partially w.r.t. to x

$\dfrac {\partial z}{\partial y} = yx^{y-1}$

Differentiating partially w.r.t. to y

$\dfrac {\partial z}{\partial y} = x^y \log x$

Now, $\delta z = \delta x \dfrac {\partial z}{\partial x} + \delta y \dfrac {\partial z}{\partial y} = \delta x (yx^{y-1}) + \delta y (x^y\log x)$

On substituting the values,

$\delta z = (0.04)(3)(1)+ (0.01) (\log (1)) = 0.12$

And $z=1^3 = 1$

Hence,

$(1.04)^{3.01} = 1 + 0.12$

$(1.04)^{3.01} = 1.12$