Let $A :=\{(x,y,z)\in\mathbb{R}^3:e^{x+y+2z}\sin(x+y)=0, e^{x+y-z}\sin(x+2y+3z)+z=0\}$. Verify that $A$ is a submanifold of dimension $1$ in $\mathbb{R}^3$ in a neighborhood of $(0,0,0)$ and then determine a base for the tangent space $TA_{(0,0,0)}$ and normale space to $A$ in the same point
Here's what I've done:
Let $f=(f_1,f_2)=(e^{x+y+2z}\sin(x+y), e^{x+y-z}\sin(x+2y+3z)+z)$; then the Jacobian matrix of the map $(f_1,f_2,z)\colon \mathbb{R}^3\to\mathbb{R}^3$ is $Jf(0,0,0)=\begin{bmatrix} 1 & 1 & 0\\ 1 & 2 & 4\\ 0 & 0 & 1 \end{bmatrix}$ and since $det(Jf(0,0,0))\neq 0$ we have that there exists a neighborhood $U_{(0,0,0)}\in\mathbb{R}^3$ such that $(U_{(0,0,0)},f_1,f_2,z)$ is a chart of $\mathbb{R}$ in $(0,0,0)$ and since in this chart the set $U_{(0,0,0)}\cap A$ is defined by the vanishing of the first two coordinates $f_1,f_2$ we have that $A$ is a submanifold of $\mathbb{R}^3$ of dimension $1$. A basis of the tangent space is a basis of $Ker(df((0,0,0)))=Ker(\begin{bmatrix} 1 & 1 & 0\\ 1 & 2 & 4 \end{bmatrix})=\{(4,4,-1)\}$ and a base for the normal space is given by $\{\nabla f_{1_{(0,0,0)}}, \nabla f_{2_{(0,0,0)}}\}=\{(1,1,1), (1,2,4)\}$.
Is this right?
Best regards,
lorenzo