Using differentials find approximately $h(1.02,1.99)$ using that
$$h=f\circ\vec g,\quad f(u,v)=3u+v^2,\quad\vec g(1,2)=(3,6),\quad D_{\vec g}(1,2)=\left(\begin{matrix}2&1\\3&5\end{matrix}\right).$$
What does it mean by "Using differentials"?
Anyway let $h(x,y)$ a differentiable function, because both $f$ and $\vec g$ are differentiable because the first one is a composition of sums and cuadratic, and the second because that has given us the jacobian, therefore their composition is also differentiable $\color{red}{(\text{in only one point or in the domain of }h)}$? So we must find $$h(x,y)=h(x_0,y_0)+h'_x(x_0,y_0)(x-x_0)+h'_y(x_0,y_0)(y-y_0).$$
Using the composition of functions I get
$$\begin{cases}h'_x&=f'_u\cdot g'_x+f'_v\cdot g'_x\\h'_y&=f'_u\cdot g'_y+f'_v\cdot g'_y,\end{cases}$$ but I am not able to continue because I don't know how to evaluate the derivatives of $\vec h$ (for $f$ yes, finding its gradient). Also I do not know what is the value of $h(x_0,y_0)$.
Can anyone help me, please?
Thank you!
Supposing that $$D_{\vec g}(1,2)=\left(\begin{matrix}2&1\\3&5\end{matrix}\right)$$ you also have $D_f(3,6)= (3,12)$.
Now you can write $(1.02,1.99)=(1+0.02,2.0-0.01)=(1,2)+(0.02,-0.01)=(1,2)+(h,k)$.
Based on that, you can use the chain rule (which you used in your question using coordinates), namely
$$D_h(1,2)=D_f(3,6) \circ D_{\vec g}(1,2)$$
To get
$$h(1.02,1.99)\approx h(1,2)+D_h(1,2).(h,k)=45+(3,12)\left[\begin{pmatrix}2 &1\\ 3&5\end{pmatrix}\begin{pmatrix}0.02\\ -0.01\end{pmatrix}\right]=45.21$$