I have a function $$\vec{y} = F(\vec{p},\vec{x})$$ where $\vec{x}$ and $\vec{y}$ $\in \mathbb{R}^6$ and $\vec{p} \in \mathbb{R}^3$.
Derivatives of $F$ are not known but there are relationships inside $\vec{y}$:
$$p_1 y_1 = y_2 y_4$$ $$p_2 y_1 y_3 = y_5$$ $$p_3 y_1 y_4 = y_6$$
And I want to determine all the partial derivatives $\frac{\displaystyle\partial{y_i}}{\displaystyle\partial{p_j}}$.
My attempt:
For y's, which are in direct relationship with $p_j$ i rewrite the condition to explicit form, eg. $$y_2 = \frac{\displaystyle p_1 y_1}{\displaystyle y_4}$$ thus $$\frac{\displaystyle\partial{y_2}}{\displaystyle\partial{p_1}} = \frac{\displaystyle y_1}{\displaystyle y_4}$$
and when there is no direct relationship, i combine two or more conditions: $$y_1 = \frac{\displaystyle y_6}{\displaystyle p_3 y_4}$$
$$\frac{\displaystyle\partial{y_2}}{\displaystyle\partial{p_3}} = \frac{\displaystyle\partial}{\displaystyle\partial{p_3}}(\frac{\displaystyle p_1 y_1}{\displaystyle y_4}) = \frac{\displaystyle\partial}{\displaystyle\partial{p_3}}(\frac{\displaystyle p_1 y_6}{\displaystyle p_3 y_4^2}) = -\frac{\displaystyle p_1 y_6}{\displaystyle p_3^2 y_4^2}$$
Is this correct? I don't feel sure with this. Please, correct me if there is any mistake.
I guess it is more convinient to use total differentiation such as $$p_1 y_1 - y_2 y_4=0 \Rightarrow dp_1y_1+p_1dy_1-dy_2y_4-y_2dy_4=0$$ Taking a partial derivative means that you have to take all other variables as fixed. In above example if you fix $y_2$ and $y_4$ it follows that $dy_2=0$ and $dy_4=0$ $$\Rightarrow dp_1y_1+p_1dy_1-0\,y_4-y_2\,0=0\Rightarrow \frac{\partial y_1}{\partial p_1}=-\frac{y_1}{p_1}$$ $$\text{by fixing $y_1$ and $y_4$}\Rightarrow \frac{\partial y_2}{\partial p_1}=\frac{y_1}{y_4}$$ $$\text{by fixing $y_1$ and $y_2$}\Rightarrow \frac{\partial y_4}{\partial p_1}=\frac{y_1}{y_2}$$ And by repeating this steps in all relations you can eliminate (or find explicit solutions for) all the partial derivatives wrt $p_1$, $p_2$ and $p_3$.