What does "class classname* funcname(void) "mean in C++?

This is an elaborated type specifier:

Elaborated type specifiers may be used to refer to a previously-declared class name (class, struct, or union) or to a previously-declared enum name even if the name was hidden by a non-type declaration. They may also be used to declare new class names.

https://en.cppreference.com/w/cpp/language/elaborated_type_specifier

Taking from answers of Artefacto and dfrib because it brings it on point: It is equivalent to:

class BOOT;
BOOT* boot(void);

In your example it essentially does a forward declaration of the class BOOT if it is not known yet. See this example struct Data* Data; from the same page:

struct Node {
    struct Node* Next; // OK: lookup of Node finds the injected-class-name
    struct Data* Data; // OK: declares type Data at global scope
                       // and also declares the data member Data
    friend class ::List; // error: cannot introduce a qualified name
    enum Kind* kind; // error: cannot introduce an enum
};
 
Data* p; // OK: struct Data has been declared

It is the same as this:

class BOOT;
BOOT* boot(void);

So it's a pointer to class BOOT, but with a declaration of the class as well. The class need not be defined at this point.

What does “class classname* funcname(void) ”mean in C++?

It is a function declaration.

It looks like a declaration of a function, but it begins with "class".

class classname* is the return type of the function. class classname is an elaborated type specifier.