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When dealing with limits involving two or more variables, it is important to understand that we are concerned with how a function behaves as its variables approach certain values simultaneously. Unlike single-variable limits where we simply approach a point from two directions (left and right), in multivariable calculus, we approach from infinitely many directions.

In the case of our problem, we need to find the limit of the function \(\frac{e^{x}}{x^{2}+y^{2}}\) as \((x, y)\) approaches \((-3,0)\). This requires checking if the limit exists irrespective of the path we take to approach the point. Commonly, we check along the main axes, and sometimes along diagonal lines such as \(y = mx\). If all paths result in the same limit, we confirm its existence.

It's crucial to recognize that path dependency reflects whether a multivariable limit exists. In our problem, the simplification gives us confidence that a specific path yields a consistent result.

L'Hopital's Rule is a powerful tool in calculus used to evaluate limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When faced with such indeterminate forms, this rule suggests that we can take the derivative of the numerator and the denominator separately and then calculate the limit of the new function.

In single-variable calculus, L'Hopital's Rule is straightforward. However, when dealing with multiple variables, we focus on one variable at a time while holding others constant at their limit value. For the problem provided, setting \(y = 0\) reduces the multivariable function to a single variable function \(\frac{e^x}{x^2}\). Upon substituting \(x = -3\), the expression becomes determinate, rendering L'Hopital's Rule unnecessary since \(\frac{e^{-3}}{9}\) is a fixed value.

Using L'Hopital's Rule requires checking the form after substitution. Only if the form is indeterminate, do we proceed with taking derivatives.

A determinate form occurs when substituting variables in a function directly yields a specific numerical value, rather than one of the indeterminate forms. In calculus, finding a determinate form means we have found the exact limit without needing extra techniques.

In the original exercise, the limit evaluation resulted in a determinate form of \(\frac{e^{-3}}{9}\). This happens when \(x = -3\) is substituted directly into the simplified form of the function, \(\frac{e^x}{x^2}\), and neither the numerator nor the denominator equals zero simultaneously.

Identifying determinate forms can save significant computation time, as it simplifies the problem to basic arithmetic. It's a reminder that while advanced tools like L'Hopital's Rule are invaluable, sometimes the simplest approach is sufficient.