I'm trying to do propagation of error using the linearized variance method (assuming independent variables, thus no need for the covariance terms):

$$\sigma^2_f = \sum^n_{k=0} \left(\frac{\partial f}{\partial x_k}\right)^2 \sigma^2_{x_k}$$

However, I have a nasty function that doesn't give me a clear-cut explicit definition of my variable. For simplicity, I will just give a simple example of what I'm trying to accomplish. Take the equation

$$x - y = e^f + e^{2f} + e^{2x}$$

This is algebraicly impossible to solving explicitly for $f = f(x,y)$. So, if I wanted to find the variance of f, I had two ideas (one of them backfired...). First, make a new function equal to zero,

$$g(x,y,f) = e^f + e^{2f} + e^{2x} + y - x = 0$$

That way I could easily find the partial derivatives, and the variance of this new function would be zero, since it's value always equaled zero.

$$\sigma^2_g = 0$$

Unfortunately, this backfired on me (after I did the 26 partial derivatives, ouch) as you can see with

$$\sigma^2_g = \left(\frac{\partial g}{\partial x}\right)^2 \sigma^2_x + \left(\frac{\partial g}{\partial y}\right)^2 \sigma^2_y + \left(\frac{\partial g}{\partial f}\right)^2 \sigma^2_f$$

If you set the variance of g to zero, then you could solve for the variance of f, and be a happy camper! Wrong. Because all the terms are squared, there is NO WAY they can add up to be zero unless they are all identically zero. That really messed me up.

The other idea I had just take the original equation and performing the partial differential with respect to x and y to each side of the equation, then solve for the partial diff quantities. That would require me to do a complete overhaul on all my work.

My question then: is there anyway to use the first method I thought of, just modifying my steps? Or maybe a third way? If not, then I will be surprised, since mathematics usually has a way to solve such twisted scenarios. Please advise soon, as I need to finish this up quickly.