$f\(1.03\,0.96$

$\doteq f\left(1\,1\right)\+\mathrm{df}$

$\=e^2\+\left(f_x\left(1\,1\right)\cdot \left(.03\right)\+f_y\left(1\,1\right)\cdot \left(-.04\right)\right)$

$\=e^2\+\left(4e^2\cdot \left(.03\right)\+4e^2\cdot \left(-.04\right)\right)$

$\=0.96 e^2$

$\=7.093493855$

Define $f\left(x\,y\right)$ and its first partial derivatives

$f\left(x\,y\right)\=x^2{\ⅇ}^{2 x y^2}$$\stackrel{\text{assign as function}}{\to }$$f$

$\mathrm{f\_\_x}\left(x\,y\right)\=\frac{\partial }{\partial x} f\left(x\,y\right)$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_x}$

$\mathrm{f\_\_y}\left(x\,y\right)\=\frac{\partial }{\partial y} f\left(x\,y\right)$$\stackrel{\text{assign as function}}{\to }$$\mathrm{f\_\_y}$

Calculate $f\left(1.03\,0.96\right)\doteq f\left(1\,1\right)\+\mathrm{df}$ 

$f\left(1\,1\right)\+\left(\mathrm{f\_\_x}\left(1\,1\right)\cdot \left(.03\right)\+\mathrm{f\_\_y}\left(1\,1\right)\cdot \left(-.04\right)\right)$ = $0.96{\ⅇ}^2$$\stackrel{\text{at 10 digits}}{\to }$$7.093493855$$$

Compute $f\left(1.03\,0.96\right)$ "exactly"

$f\left(1.03\,0.96\right)$ = $7.082405462$$$

Define an appropriate function $f$ 

$f≔\left(x\,y\right)\to x^2{\ⅇ}^{2 x y^2}\:$

Compute $\mathrm{df}\=f_x\left(1\,1\right)\cdot \left(0.03\right)\+f_y\left(1\,1\right)\cdot \left(-0.04\right)$ 

$\mathrm{df}≔\mathrm{evalf}\left({\mathrm{D}}_1\left(f\right)\left(1\,1\right)\cdot \left(.03\right)\+{\mathrm{D}}_2\left(f\right)\left(1\,1\right)\cdot \left(-.04\right)\right)$ = $-0.2955622440$$$

Approximate $f\left(1.03\,0.96\right)$ as $f\left(1\,1\right)\+\mathrm{df}$ 

$\mathrm{evalf}\left(f\left(1\,1\right)\+\mathrm{df}\right)$ = $7.093493855$$$

Obtain $f\left(1.03\,0.96\right)$ "exactly"

$f\left(1.03\,0.96\right)$ = $7.082405462$$$