The multivariable chain rule when moving along a curve If I have some curve moving through $\Bbb R^2$ parameterized by $\lambda$, $R(\lambda)$, I can expand an infinitesimal movement along the curve using the multivariable chain rule:
$$\frac{dR}{d\lambda}=\frac{\partial R}{\partial x}\frac{dx}{d\lambda}+\frac{\partial R}{\partial y}\frac{dy}{d\lambda},$$
which I understand geometrically for the case of a scalar function, you move along the x-axis slightly and multiply the distance by the gradient of the function along the x-axis ($\frac{\partial R}{\partial x})$ and add it to the same procedure in the y-direction.
However with a curve moving through $\Bbb R^2$ you don't seem to be able to split the movement up into an x followed by a y movement since any change in x will involve some change in y and visa versa.
My question is essentially how does this work?
Poorly drawn visualisation:

 A: It is not clear what $R$ is in your question, but suppose $R$ is a function of three variables and a curve is given for each value of $\lambda$ by the equation $R(x,y,\lambda) = 0$. For example, circles $x^2+y^2 -\lambda^2  = 0$. 
The way you wrote the chain rule would apply if $x$ and $y$ were functions of $\lambda$, but that is not the case with this interpretation of what $R$ means.
The Implicit Function Theorem says roughly that if the partial derivative $R_y$
is not zero at some point of the set where $R(x,y,\lambda) = 0$, then you can solve
locally near that point for $y$ as a function of $x$ and $\lambda$. That means it really is a set of curves. In our circle example you have $R_y = 2y \ne 0$ at the point $(x,y,\lambda) = (0,3,3)$ and you can solve near there as $y = \sqrt{\lambda^2-x^2}$. In this example "near" means the whole upper semicircles with any positive radius. 
To relate all the derivatives then you can differentiate  $R(x,y(x,\lambda),\lambda) = 0$ with respect to $x$ or $\lambda$ but not $y$, giving $R_x+R_y y_x = 0$ and $R_y y_\lambda + R_\lambda = 0$. For our circles example those read $2x+2y y_x = 0$ and $2y y_\lambda -2\lambda = 0$, which can be put into terms of the independent variables $x,\lambda$ as $y_x = -\frac{x}{\sqrt{\lambda^2-x^2}}$ and $y_\lambda = \frac{\lambda}{\sqrt{\lambda^2-x^2}}$.
For the Implicit Function Theorem see for example Multivariable Calculus With Applications by Lax and Terrell, from Springer.
