given a function $F:R^{3}\rightarrow R,$ such that $F(x,y,2x^{2}+y^{2})=3x-5y,$ define $u(x,y)=2x^{2}+y^{2}$
find the gardient at the point (1,2,6).
i am also given that $\frac{\partial F}{\partial x}(1,2,6)+2\frac{\partial F}{\partial y}(1,2,6)+2\frac{\partial F}{\partial u}(1,2,6)=\sqrt{5}.$
my solution: define $g:R^{2}\rightarrow R,g(x,y)=F(x,y,u(x,y)),$whice means $g(x,y)=3x-5y.$
there for: $\frac{\partial g}{\partial x}(1,2,6)=\frac{\partial F}{\partial x}(1,2,6)\cdot1+\frac{\partial F}{\partial y}(1,2,6)\cdot0+\frac{\partial F}{\partial u}(1,2,6)\cdot\frac{du}{dx}(1,2,6)$.
the RHS euqlas $\frac{\partial F}{\partial x}(1,2,6)+4\cdot\frac{\partial F}{\partial u}(1,2,6)$ while the LHS is $3$. so i get:
$\frac{\partial F}{\partial x}(1,2,6)+4\cdot\frac{\partial F}{\partial u}(1,2,6)=3$
and then i can do similarly for $y$ and get 3 equations with 3 variables? is the eqation i got here correct?
Write $$ v(x,y)=\begin{pmatrix}x\\y\\u(x,y)\end{pmatrix}\,. $$ We know from the chain rule that $$\tag{1} \nabla (F\circ v)=J^\top(v)\,\nabla F $$ where $J^\top(v)$ is the transpose Jacobian of $v\,:$ $$ J^\top(v)=\begin{pmatrix}1 & 0 & \partial_x u\\ 0 & 1 & \partial_y u\end{pmatrix}= \begin{pmatrix}1 & 0 & 4x\\ 0 & 1 & 2y\end{pmatrix}\,. $$
The LHS of (1) is $$ \nabla (3x-5y)=\begin{pmatrix}3\\-5\end{pmatrix}\,. $$ Therefore the gradient of $F$ must satisfy \begin{align} 3&=\partial_xF+4\,x\,\partial_zF\\[2mm] -5&=\partial_yF+2\,y\,\partial_zF\,. \end{align} At $(1,2,6)$ this is \begin{align} 3&=\partial_xF+4\,\partial_zF\\[2mm] -5&=\partial_yF+4\,\partial_zF\,. \end{align} The other equation you are given I write as $$ \partial_x F+2\,\partial_y\,F+2\,\partial_z F=\sqrt{5}\,,\quad (x,y,z)=(1,2,6)\,. $$ You did not solve the three equations with the three unknowns $\partial_xF,\partial_yF,\partial_zF$ that we have now.
Can you proceed ?